训练合集-Mind the Gap

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wirtten by xishuiw

训练合集-Mind the Gap
- 2023暑期训练前半
- 2023暑期训练后半

2023暑期训练前半

0.2022-2023 ACM-ICPC Latin American Regional Programming Contest

赛时过题 DEILM

补题ACH

D.Daily Trips

签到问什么时候需要带伞

#include<bits/stdc++.h>
#define close std::ios::sync_with_stdio(false),cin.tie(0),cout.tie(0)
using namespace std;
typedef long long ll;
const ll MAXN =  3e5+7;
const ll mod = 1e9+7;
const ll inf = 0x3f3f3f3f;
void solve(){
	int n,a,b;cin>>n>>a>>b;
	for(int i=1;i<=n;i++){
		char ch1,ch2;
		cin>>ch1>>ch2;
		if(ch1=='Y'||b==0){
			cout<<"Y ";
			a--;
			b++;
		}
		else {
			cout<<"N ";
		}
		if(ch2=='Y'||a==0){
			cout<<"Y ";
			b--;
			a++;
		}
		else {
			cout<<"N ";
		}
		cout<<"\n";
	} 
}
signed main(){
	close;
	solve();
}

E.Empty Square

#include <iostream>
using namespace std;

const int N = 1007;
int vis[N];

int main()
{
    std::ios::sync_with_stdio(false);
    cin.tie(0);
    cout.tie(0);
    int n, k, e;
    cin >> n >> k >> e;
    vis[k] = 1;
    int l = e, r = n - k - e;
    int ans = 0;
    if (l > r)
        swap(l, r);
    if (l == 1)
    {
        if (vis[1])
            ans++;
        else
            vis[1] = 1;
    }
    else if (l == 2)
    {
        if (vis[2])
        {
            vis[1] = 1;
            ans++;
        }
        else
            vis[2] = 1;
    }
    else if (l == 3)
    {
        if (vis[3])
        {
            vis[1] = vis[2] = 1;
        }
        else
        {
            vis[3] = 1;
        }
    }
    else if (l == 4)
    {
        if (vis[4] == 1)
        {
            vis[1] = 1;
            vis[3] = 1;
        }
        else
        {
            vis[4] = 1;
        }
    }

    if (r == 1)
    {
        if (vis[1])
            ans++;
    }
    else if (r == 2)
    {
        if (vis[2])
        {
            if (vis[1])
                ans += 2;
            else
                ans++;
        }
    }
    else if (r == 3)
    {
        if (vis[3])
        {
            if (vis[2])
            {
                if (vis[1])
                {
                    ans += 3;
                }
                else
                    ans += 2;
            }
            else
            {
                if (vis[1])
                {
                    ans += 1;
                }
            }
        }
    }
    else if (r == 4)
    {
        if (vis[4] == 1)
        {
            if (vis[3])
            {
                if (vis[2])
                {
                    if (vis[1])
                        ans += 4;
                    else
                        ans += 3;
                }
                else
                {
                    if (vis[1])
                        ans += 2;
                    else
                        ans++;
                }
            }
            else
            {
                if (vis[1])
                    ans++;
            }
        }
    }
    cout << ans << endl;
}

I. Italian Calzone & Pasta Corner

找一个最大上升子序列路径里图很小对每个点跑bfs 注意剪枝

#include<bits/stdc++.h>
#define close std::ios::sync_with_stdio(false),cin.tie(0),cout.tie(0)
using namespace std;
typedef long long ll;
const ll MAXN =  3e5+7;
const ll mod = 1e9+7;
const ll inf = 0x3f3f3f3f;
int mp[300][300];
int vis[300][300],f[300][300];
int fx[4][2]={1,0,-1,0,0,1,0,-1};
void solve(){
	int n,m;cin>>n>>m;
	 for(int i=1;i<=n;i++){
	 	for(int j=1;j<=m;j++) cin>>mp[i][j];
	 }
	 int ans=0;
	 for(int i=1;i<=n;i++){
	 	for(int j=1;j<=m;j++){
	 		int res=0;
	 		if(vis[i][j]) continue;
	 		for(int x=1;x<=n;x++){
	 			for(int y=1;y<=m;y++) f[x][y]=0;
			 }
			 queue<array<int,2>> que;
			 que.push({i,j});
			 f[i][j]=1;
			 vis[i][j]=1;
			 res=1;
			 while(que.size()){
			 	int x=que.front()[0],y=que.front()[1];
			 	que.pop();
			 	for(int v=0;v<4;v++){
			 		int xx=x+fx[v][0],yy=y+fx[v][1];
			 		if(xx<1||xx>n||yy<1||yy>m||f[xx][yy]||mp[x][y]>mp[xx][yy]) continue;
			 		f[xx][yy]=1;res++; 
			 		vis[xx][yy]=1;
			 		que.push({xx,yy});
				 }
			 	
			 }
			 ans=max(ans,res);
		 }
	 }
	 cout<<ans<<"\n";
}
signed main(){
    close;
	solve();
}

L. Lazy Printing

给打印机找长度在限制范围内的最小的模板串数量

贪心地不停向后枚举然后用hash优化下过了但是这个还是n2的数据太弱标准做法kmp

#include <bits/stdc++.h>

#define close                         \
	std::ios::sync_with_stdio(false); \
	cin.tie(0);                       \
	cout.tie(0)
using namespace std;
typedef unsigned long long ULL;

const int N = 1007;
const int MANX = 4e5+7;
int n, d,P=131;
char s[200010];
const int inf = 2e9;
ULL h[MANX], p[MANX];
ULL get(int l, int r) {
	return h[r] - h[l - 1] * p[r - l + 1];
}
inline int find(int now, int i) {
//	int pos = now;
	int t = 0;
	int ans=now;
	int pos=now;
	pos+=i;
	ULL k=get(now,pos);
	while(pos<=n){
		ULL tmp=get(pos-i,pos);
		if(tmp!=k) break;
		ans=pos;
		pos+=i;
	}
	pos=now;
	while (s[pos] == s[ans]) {
		++pos;
		++ans;
		++t;
		if (t == i) {
			t = 0;
			pos = now;
		}
	}
	return ans;
}
int main() {
	close;
	p[0] = 1;
	cin >> s+1 >> d;
	n = strlen(s+1);
	for (int i = 1; i <= n; i ++ ) {
		h[i] = h[i - 1] * P + s[i];
		p[i] = p[i - 1] * P;
	}

	int now = 1;
	int ans = 0;
	while (now <= n) {
		int mx = 0;
		for (int i = 1; i <= d; ++i) {
			int k = find(now, i);
			if (k > mx) {
				mx = k;
			}
		}
		++ans;
		now = mx;
	}
	cout << ans;
}

M. Maze in Bolt

可以旋转的锁一步一步转看看能不能下去并且可以换头的方向

队友搓的搜索很牛

#include <iostream>
#include <queue>
using namespace std;

const int N = 1007;
char mp[N][N];
int row, col;
int flag[N][N];
int vis[N][N];
char tag[N];
int bg = -1;
int dx[4] = {0, 0, 1, -1};
int dy[4] = {1, -1, 0, 0};

bool check(int r, int c)
{
    for (int i = 0; i < col; i++)
    {
        if (tag[(bg + i) % col] == '0')
            continue;
        if (mp[r][(c + i) % col] == '1')
            return false;
    }
    return true;
}

bool bfs()
{
    for (int i = 1; i <= row; i++)
    {
        for (int j = 0; j < col; j++)
        {
            if (check(i, j))
                flag[i][j] = 1;
            else
                flag[i][j] = 0;
            vis[i][j] = 0;
            // if (flag[i][j])
            //     cout << i << ' ' << j + 1 << endl;
        }
    }
    queue<pair<int, int>> q;
    bool getans = false;
    for (int i = 0; i < col; i++)
    {
        if (flag[1][i])
        {
            vis[1][i] = 1;
            q.push({1, i});
        }
    }
    while (q.size())
    {
        int x = q.front().first, y = q.front().second;
        q.pop();
        if (x == row)
        {
            getans = true;
            break;
        }
        for (int i = 0; i < 4; i++)
        {
            int xx = x + dx[i];
            int yy = (y + dy[i] + col) % col;
            if (xx <= 0 || xx > row)
                continue;
            if (vis[xx][yy])
                continue;
            if (flag[xx][yy] == 0)
                continue;
            vis[xx][yy] = 1;
            q.push({xx, yy});
        }
    }
    return getans;
}

int main()
{
    std::ios::sync_with_stdio(false);
    cin.tie(0);
    cout.tie(0);
    cin >> row >> col;
    for (int i = 0; i < col; i++)
    {
        cin >> tag[i];
        // cout << tag[i];
        if (tag[i] == '1' && bg == -1)
            bg = i;
    }
    // cout << endl;
    // cout << bg << endl;
    for (int i = 1; i <= row; i++)
    {
        for (int j = 0; j < col; j++)
        {
            cin >> mp[i][j];
        }
    }
    if (bg == -1)
    {
        cout << "Y";
        return 0;
    }
    if (bfs())
        cout << "Y";
    else
    {
        for (int i = 0; i < col - i - 1; i++)
        {
            // cout << i << ' ' << col - i << endl;
            swap(tag[i], tag[col - i - 1]);
        }
        for (int i = 0; i < col; i++)
        {
            // cout << tag[i];
            if (tag[i] == '1')
            {
                bg = i;
                break;
            }
        }
        // cout << endl;
        if (bfs())
            cout << "Y";
        else
            cout << "N" << endl;
    }
}

A. Asking for Money

每个人被一个人要钱时候可以对两个人要钱

随机某个人被要1元问谁最后有可能不亏钱

不亏钱条件是有一个点能同时到达三个点

建反图三个点同时到达一个点复杂度大大优化

#include<bits/stdc++.h>
#define close std::ios::sync_with_stdio(false),cin.tie(0),cout.tie(0)
using namespace std;
typedef long long ll;
const ll MAXN =  3e5+7;
const ll mod = 1e9+7;
const ll inf = 0x3f3f3f3f;
vector<int> adj[MAXN];
pair<int,int> e[MAXN];
int vis[MAXN],tmp[MAXN];
void solve(){
	int n;cin>>n;
	for(int i=1;i<=n;i++){
		int u,v;cin>>u>>v;
		adj[u].push_back(i);
		adj[v].push_back(i);
		e[i]={u,v};
	}
	for(int i=1;i<=n;i++){
		for(int it=1;it<=n;it++) vis[it]=0,tmp[it]=0;
		queue<int>que;
		que.push(i);
		vis[i]=1;
		while(que.size()){
			int x=que.front();
			que.pop();
			for(auto &v:adj[x]){
				if(!vis[v]){
					vis[v]=1;
					que.push(v);
				}
			}
		}
		for(int j=1;j<=n;j++) if(vis[j]) tmp[j]++;
		//
		for(int it=1;it<=n;it++) vis[it]=0;
		que.push(e[i].first);
		vis[e[i].first]=1;
		while(que.size()){
			int x=que.front();
			que.pop();
			for(auto &v:adj[x]){
			    if(v==i) continue;
				if(!vis[v]){
					vis[v]=1;
					que.push(v);
				}
			}
		}
		for(int j=1;j<=n;j++) if(vis[j]) tmp[j]++;
		//
		for(int it=1;it<=n;it++) vis[it]=0;
		que.push(e[i].second);
		vis[e[i].second]=1;
		while(que.size()){
			int x=que.front();
			que.pop();
			for(auto &v:adj[x]){
			    if(v==i) continue;
				if(!vis[v]){
					vis[v]=1;
					que.push(v);
				}
			}
		}
		for(int j=1;j<=n;j++) if(vis[j]) tmp[j]++;
		int flag=0;
		for(int j=1;j<=n;j++){
			if(tmp[j]==3) flag=1;
		}
		if(flag) cout<<"Y";
		else cout<<"N";
	}
}
signed main(){
	solve();
}

C. City Folding

码力太弱推不出公式

#include<bits/stdc++.h>
#define close std::ios::sync_with_stdio(false),cin.tie(0),cout.tie(0)
using namespace std;
typedef long long ll;
const ll MAXN =  3e5+7;
const ll mod = 1e9+7;
const ll inf = 0x3f3f3f3f;
#define int ll
int f[MAXN];
vector<int> ans;
void solve(){
	int n,p,h;cin>>n>>p>>h;
	p--;h--;
	for(int i=n-1;i>=0;i--){
		if(h>=(1LL<<i)) {
			f[n-i-1]=1;
			h=(1LL<<(i+1))-1-h;
		}
	//cout<<"# h = "<<h<<" "<<f[n-i-1]<<"\n";
	} 
	//预处理是否需要翻动来增加高度，f为1表示需要增加一次高度
	//3 8 3 第一次翻动增加4 不需要 第二次翻动增加2 需要，翻动后长度变为剩下的长度 
	for(int i=n-1;i>=0;i--){
		//第i次翻动 h改变的是1<<(n-i-1) p改变的是1<<i 
		if(!f[i]){ 
	//	cout<<f[i];//需要增加h 
			if(p<(1LL<<i)){
				cout<<"R"; 
			}
			else{
				cout<<"L";
				p-=(1LL<<i);
			}
		}
		else{//需要的长度大 
	//	cout<<f[i];
		//反转后高度会改变 
			if(p<(1LL<<i)){
				cout<<"L";
				p=(1LL<<i)-1-p; 
			}
			else{
				cout<<"R";
				p=(1LL<<(i+1))-1-p;
			}
		}
	} 
}
signed main(){
	solve();
}

1.The 2022 ICPC Asia Nanjing Regional Contest

赛时过题AGIM 补题D

A. Stop, Yesterday Please No More

有很多袋鼠满的然后会根据题目给的走问有几种放洞的方法可以让最后剩下k个袋鼠

我一开始在走洞写了一个自己样例都过不去的做法

队友上机敲了一发走袋鼠的直接过了

#include <iostream>
#include <vector>
#include <queue>
#include <algorithm>
#include <string>
#include <stack>
#include <iomanip>
// 小数位    cout<<fixed<<setprecision(12)<<dis[n+1]<<endl;
using namespace std;
#define close                         \
    std::ios::sync_with_stdio(false); \
    cin.tie(0);                       \
    cout.tie(0)

const int N = 1e3 + 7;
typedef long long ll;

int row, col, lv;
string s;
ll vis[N][N];

ll count(int x1, int y1, int x2, int y2)
{
    // cout << vis[x2][y2] << " " << vis[x1 - 1][y1 - 1] << ' ' << vis[x1 - 1][y2] << ' ' << vis[x2][y1 - 1] << endl;
    return vis[x2][y2] + vis[x1 - 1][y1 - 1] - vis[x1 - 1][y2] - vis[x2][y1 - 1];
}

void solve()
{
    cin >> row >> col >> lv;
    for (int i = 1; i <= row; i++)
        for (int j = 1; j <= col; j++)
            vis[i][j] = 0;
    cin >> s;
    ll v = 0, h = 0; // 竖直，水平
    ll mxv0, mxv1, mxh0, mxh1;
    mxv0 = mxv1 = mxh0 = mxh1 = 0;
    ll x = 1, y = 1;
    for (int i = 0; i < s.length(); i++)
    {
        if (s[i] == 'U')
        {
            v++;
            x = max(x, v + 1);
            mxv0 = max(mxv0, v);
        }
        else if (s[i] == 'D')
        {
            v--;
            mxv1 = min(mxv1, v);
        }
        else if (s[i] == 'L')
        {
            h--;
            y = max(y, -h + 1);
            mxh1 = min(mxh1, h);
        }
        else
        {
            h++;
            mxh0 = max(mxh0, h);
        }
    }

    ll szrow = row - mxv0 + mxv1, szcol = col - mxh0 + mxh1;
    if (szrow <= 0 || szcol <= 0)
    {
        if (lv == 0)
            cout << row * col << endl;
        else
            cout << 0 << endl;
        return;
    }

    // cout << szrow << ' ' << szcol << endl;
    // cout << x << ' ' << y << endl;
    vis[x][y] = 1;
    for (int i = 0; i < s.length(); i++)
    {
        if (s[i] == 'U')
        {
            x--;
        }
        else if (s[i] == 'D')
        {
            x++;
        }
        else if (s[i] == 'L')
        {
            y--;
        }
        else
        {
            y++;
        }
        vis[x][y] = 1;
    }
    for (int i = 1; i <= row; i++)
    {
        for (int j = 1; j <= col; j++)
        {
            vis[i][j] += -vis[i - 1][j - 1] + vis[i - 1][j] + vis[i][j - 1];
        }
    }
    // for (int i = 1; i <= row; i++)
    // {
    //     for (int j = 1; j <= col; j++)
    //     {
    // cout << vis[i][j] << ' ';
    // }
    // cout << endl;
    // }
    ll ans = 0;
    // cout << count(2, 2, 3, 3) << endl;
    ll sz = szrow * szcol;

    // for (int i = szrow; i <= row; i++)
    // {
    //     for (int j = szcol; j <= col; j++)
    //     {
    //         ll cnt = count(i - szrow + 1, j - szcol + 1, i, j);
    //         if (cnt + lv == sz)
    //             ans++;
    //     }
    // }
    // cout << ans << ' ';
    // ans = 0;
    for (int i = 1; i <= row; i++)
    {
        for (int j = 1; j <= col; j++)
        {
            ll cnt = count(max(1ll, i - szrow + 1), max(1ll, j - szcol + 1), i, j);
            if (cnt + lv == sz)
                ans++;
        }
    }
    cout << ans << endl;
}

int main()
{
    close;
    int t = 1;
    cin >> t;
    while (t--)
    {
        solve();
    }
}

G. Inscryption

野兽合并写太急了被我wa了一发

显然是合并越多越好 0可以变成1 所以是类似反悔贪心

0多了就把1个0变1

#include<bits/stdc++.h>
#define close std::ios::sync_with_stdio(false),cin.tie(0),cout.tie(0)
using namespace std;
typedef long long ll;
#define int ll
const ll MAXN =  1e6+7;
const ll mod = 1e9+7;
const ll inf = 0x3f3f3f3f;
int a[MAXN];
void solve(){
	int n;cin>>n;
	int cnt0=0,cnt1=0,cnt2=0;
	int now=0;
	int flag=1;
	for(int i=1;i<=n;i++){
		cin>>a[i];
		if(a[i]==0) cnt0++;
		else if(a[i]==1) cnt1++;
		else cnt2++;
		if(a[i]!=0)
		now+=a[i];
		else now-=1;
		if(now<0){
			if(cnt0){
				cnt0--;
				cnt1++;
				now+=2;
			}
			else flag=0;
		}
	}

	int fz=1+cnt1,fm=1+cnt1-cnt0-cnt2;
	if(!flag) cout<<"-1\n";
	else{
		int tmp=__gcd(fz,fm);
		cout<<fz/tmp<<" "<<fm/tmp<<"\n"; 
	}
}
signed main(){
	close; 
	int t;cin>>t;
	while(t--) 
	solve();
}

I. Perfect Palindrome

结论：全变成一样的

#include<bits/stdc++.h>
#define ll long long
#define PII pair<int,int>
//  #define int long long
using namespace std;
constexpr ll mod=1e9+7; 
const ll inf=0x3f3f3f3f;  
const ll INF=0x3f3f3f3f3f3f3f3f;  
const double eps=1e-10;  
const int N=2e5+10;
// struct node{
//     friend bool operator<(const node&a,const node&b){
//         return ;
//     }
// }
//priority_queue<ll,vector<ll>,greater<ll>>pq;
inline int read(){int x=0,f=1;char ch=getchar();while(ch<'0'||ch>'9'){if(ch=='-')f=-1;ch=getchar();}while(ch>='0'&&ch<='9'){x=(x<<1)+(x<<3)+(ch^48);ch=getchar();}return x*f;}
inline void write(int x){char F[200];int tmp=x>0?x:-x;if(x<0) putchar('-');int cnt=0;while(tmp>0){F[cnt++]=tmp%10+'0';tmp/=10;}while(cnt>0)putchar(F[--cnt]);}
inline int combination(int n,int k){int sum=0;if (n==k||k==0){return 1;}else{return combination(n-1,k)+combination(n-1,k-1);}}
map<char,int>mp;
void solve(){
    // int n=read();
    string s;
    mp.clear();
    cin>>s;
    for(int i=0;i<s.size();i++){
        mp[s[i]]++;
    }
    int maxx=-1;
    for(auto it:mp){
        int y=it.second;
        maxx=max(y,maxx);
    }
    cout<<s.size()-maxx<<'\n';
    //puts(ans>0?"YES":"NO");
    //puts(ans>0?"Yes":"No");
}

signed main(){
    // ios::sync_with_stdio(false);
    // cin.tie(0);
    // cout.tie(0);
    // int t=1;
   int t=read();
    while(t--){
        solve();
    }
}

M. Drain the Water Tank

队友写的向量判上边和下边因为给定顺序排过了所以就看看是不是下凹和下平

#include <iostream>
#include <vector>
#include <queue>
#include <algorithm>
#include <string>
#include <stack>
#include <iomanip>
// 小数位    cout<<fixed<<setprecision(12)<<dis[n+1]<<endl;
using namespace std;
#define close                         \
    std::ios::sync_with_stdio(false); \
    cin.tie(0);                       \
    cout.tie(0)

const int N = 2e5 + 7;
const int inf = 2e9;

struct node
{
    int x, y;
} nd[N];

int n;

void solve()
{
    cin >> n;
    int mn = inf, index = 0;
    for (int i = 0; i < n; i++)
        cin >> nd[i].x >> nd[i].y;

    int ans = 0;
    for (int i = 0; i < n; i++)
    {
        int x = nd[i].x, y = nd[i].y;
        int nx = (i + 1) % n;
        int xx = nd[nx].x, yy = nd[nx].y;
        int nxx = (nx + 1) % n;
        int xxx = nd[nxx].x, yyy = nd[nxx].y;

        if (x < xx && y > yy)
        {
            if (yyy > yy && (xxx >= xx || (yyy - yy) * (x - xx) < (y - yy) * (xxx - xx)))
                ans++;
        }
        else if (x == xx && y > yy)
        {
            if (yyy > yy && xxx > xx)
                ans++;
        }
        else if (x > xx && y > yy)
        {
            if (yyy > yy && xxx > xx && (yyy - yy) * (x - xx) < (y - yy) * (xxx - xx))
                ans++;
        }
    }
    // cout << ans << endl;
    for (int i = 0; i < n; i++)
    {
        int nx = (i + 1) % n;
        while (nd[i].y == nd[nx].y && nd[i].x < nd[nx].x)
            nx = (nx + 1) % n;
        if (nx == (i + 1) % n)
            continue;
        int pre = (n + i - 1) % n;
        if (nd[pre].y > nd[i].y && nd[(n + nx - 1) % n].y < nd[nx].y)
            ans++;
    }
    cout << ans << endl;
}

int main()
{
    close;
    int t = 1;
    // cin >> t;
    while (t--)
    {
        solve();
    }
}

D. Chat Program

一次可以加一个等差数列问第k大的数最多是多大

二分答案看看加上后>=mid的数是否大于k个

#include<bits/stdc++.h>
#define close std::ios::sync_with_stdio(false),cin.tie(0),cout.tie(0)
using namespace std;
typedef long long ll;
const ll MAXN =  3e5+7;
const ll mod = 1e9+7;
const ll inf = 0x3f3f3f3f;
const ll INF = 0x3f3f3f3f3f3f3f3f;
#define int ll
int n,k,m,c,d;
int a[MAXN],pre[MAXN];
bool check(int mid){
	int cnt=0;
	for(int i=1;i<=n;i++) pre[i]=0;
	for(int i=1;i<=n;i++){
		if(a[i]>=mid) cnt++;
		else{
			int tmp=mid-a[i];
			tmp-=c;
			int pos;
			if(tmp<=0) {
				pos=max(i-m+1,1LL);
				pre[pos]++;
				pre[i+1]--;
			}
			else{
				int p;
				if(d==0){
					continue;
				}
				p=tmp/d;
				if(p*d!=tmp) p++;
				if(p>m) continue;
				pos=i-p;
				if(pos<1) continue;
				p=max(i-m+1,1LL);
				pre[p]++;
				pre[min(pos+1,i+1)]--; 
			} 
		}
	}
	pre[1]+=cnt;
	for(int i=1;i<=n;i++){
		pre[i]=pre[i]+pre[i-1];
		if(pre[i]>=k) return true;
	}
	return false;
}
void solve(){
	cin>>n>>k>>m>>c>>d;
	for(int i=1;i<=n;i++) cin>>a[i];
	int l=0,r=INF;
	while(l<=r){
		int mid=l+r>>1;
		if(check(mid)){
			l=mid+1;
		}
		else{
			r=mid-1;
		}
	} 
	cout<<r;
}
signed main(){
	close;
	solve();
}

2.2019-2020 ICPC Asia Hong Kong Regional Contest

赛时过题BDG 补题EJ

B. Binary Tree

每次拿不改变奇偶性算树的总大小就行了

#include<bits/stdc++.h>
#define close std::ios::sync_with_stdio(false),cin.tie(0),cout.tie(0)
using namespace std;
typedef long long ll;
const ll MAXN =  3e5+7;
const ll mod = 1e9+7;
const ll inf = 0x3f3f3f3f;
void solve(){
	int n;cin>>n;
	for(int i=1;i<n;i++){
		int u,v;cin>>u>>v;
	}
	if(n%2==1) cout<<"Alice\n";
	else cout<<"Bob\n";
}
signed main(){
	close;
	int t;cin>>t;
	while(t--)
	solve();
}

D. Defining Labels

类似进制转换队友写的

#include <iostream>
#include <vector>
#include <stack>
#include <map>
#include <queue>
#include <string>
#include <algorithm>
#include <iomanip> //fixed<<setprecision(12)<<
#include <utility> //pair模板
#include <cmath>
#include <random> //mt19937 mtrand(time(0));
#include <time.h>
#define close                         \
    std::ios::sync_with_stdio(false); \
    cin.tie(0);                       \
    cout.tie(0)
typedef long long ll;
typedef unsigned long long ull;
const int N = 2e5 + 2;
const ll mod = 200907;
using namespace std;

int k, x;

void solve()
{
    //	freopen("data.in.txt","r",stdin);
    //	freopen("data.out.txt","w",stdout);
    cin >> k >> x;
    int d = 9 - (10 - k) + 1;
    // cout << d << endl;
    vector<int> ans;
    while (x > 0)
    {
        ans.push_back((x - 1) % d + (10 - k));
        // cout << (x - 1) % d + (10 - k);
        // x = (x - 1) / d + 1;
        x = (x / d) - (int)(x % d == 0);
    }
    for (int i = ans.size() - 1; i >= 0; i--)
        cout << ans[i];
    cout << endl;
}

int main()
{
    close;
    int t = 1;
    cin >> t;
    while (t--)
    {
        solve();
    }
    return 0;
}

G. Game Design

#include <iostream>
#include <vector>
#include <stack>
#include <map>
#include <queue>
#include <string>
#include <algorithm>
#include <iomanip> //fixed<<setprecision(12)<<
#include <utility> //pair模板
#include <cmath>
#include <random> //mt19937 mtrand(time(0));
#include <time.h>
#define close                         \
    std::ios::sync_with_stdio(false); \
    cin.tie(0);                       \
    cout.tie(0)
typedef long long ll;
typedef unsigned long long ull;
const int N = 2e5 + 2;
const ll mod = 200907;
using namespace std;

int k, cnt;
map<int, int> wei, fa;

void dfs(int k, int f)
{
    cnt++;
    // cout << cnt << ' ' << f << endl;
    fa[cnt] = f;
    int tp = cnt;

    if (k == 1)
    {
        wei[tp] = 1;
        wei[f] += wei[tp];
        return;
    }
    k--;
    if (k % 2 == 0)
    {
        // cout << cnt << endl;
        dfs(2, tp);
        dfs(k / 2, tp);
    }
    else
    {
        dfs(k, cnt);
    }
    wei[f] += wei[tp];
}

void solve()
{
    //	freopen("data.in.txt","r",stdin);
    //	freopen("data.out.txt","w",stdout);
    cin >> k;
    if (k == 1)
    {
        cout << 2 << endl;
        cout << 1 << endl;
        cout << 1 << ' ' << 2 << endl;
        return;
    }
    // cout << k << endl;
    dfs(k, 0);
    cout << cnt << endl;
    for (int i = 2; i <= cnt; i++)
        cout << fa[i] << ' ';
    cout << endl;
    for (int i = 1; i <= cnt; i++)
        cout << wei[i] << ' ';
    cout << endl;
}

int main()
{
    close;
    int t = 1;
    // cin >> t;
    while (t--)
    {
        solve();
    }
    return 0;
}

E. Erasing Number

能处于中位数的肯定能删

三个连续大数能变成两个的对于每个数数连续的大数n2去做

J. Junior Mathematician

数位dp 模板题赛场看出来了不会写

#include<bits/stdc++.h>
#define close std::ios::sync_with_stdio(false),cin.tie(0),cout.tie(0)
using namespace std;
typedef long long ll;
const ll MAXN =  1e4+7;
const ll mod = 1e9+7;
const ll inf = 0x3f3f3f3f;
ll dp[MAXN][62][62];//pos pre 贡献
// dp[i][j][m]
//轮流枚举每一位的情况 都可以由上一位推出 复杂度为O(10n)
//pre表示前缀和 来算此时数的大小
//第三维记录f[x]-x
//f[x] 增加了 pre*x,  x增加了x*1-^n
//贡献=贡献-x*10^n+pre*x
int p[MAXN],upper[MAXN],m;
void init(int x) {
	p[1]=1;
	for(int i=2; i<=x+1; i++) {
		p[i]=p[i-1]*10;
		p[i]%=m;
	}
}
ll dfs(int pos,int pre_num,int c,int flag) {
	int max_number;
	if(pos<=0) return c==0;
	if(!flag&&dp[pos][pre_num][c]!=-1) return dp[pos][pre_num][c];
	if(flag) max_number=upper[pos];
	else max_number=9;
	ll ret=0;
	for(int i=0; i<=max_number; i++) {
//转移方程

		ret+=dfs(pos-1,(i+pre_num)%m,((c-i*p[pos]+pre_num*i)%m+m)%m,flag&&(i==max_number));
		ret=(ret+mod)%mod;
	}
	ret=(ret+mod)%mod;
	if(!flag) dp[pos][pre_num][c]=ret;
	return ret;
}

void solve() {
	string a,b;
	cin>>a>>b>>m;
	int l=max(a.length(),b.length());
	init(l);
	int len=a.length();
	a[len-1]--;
	for(int i=0; i<=l+4; i++) {
		for(int j=0; j<=m; j++) {
			for(int x=0; x<=m; x++) {
				dp[i][j][x]=-1;
			}
		}
	}
	for(int i=len-1; i>=0; i--) {
		if(a[i]>='0'&&a[i]<='9') break;
		a[i]='9',a[i-1]--;
	}
	for(int i=0; i<len; i++) {
		upper[len-i]=a[i]-'0';
	}
	//cout<<"## "<<a<<"\n";
	ll aa=dfs(len,0,0,1);
	for(int i=0; i<=l+4; i++) {
		for(int j=0; j<=m; j++) {
			for(int x=0; x<=m; x++) {
				dp[i][j][x]=-1;
			}
		}
	}
	len=b.length();
	for(int i=0; i<len; i++) {
		upper[len-i]=b[i]-'0';
	}
	ll bb=dfs(len,0,0,1);
	cout<<(bb-aa+mod)%mod<<"\n";
}
signed main() {
	close;
	int t;cin>>t;
	while(t--) 
	solve();
}

3.The 2021 ICPC Asia Nanjing Regional Contest (XXII Open Cup, Grand Prix of Nanjing)

赛时过题ACHM 补题EJ

A. Oops, It’s Yesterday Twice More

#include <iostream>
#include <vector>
#include <stack>
#include <map>
#include <set>
#include <queue>
#include <string>
#include <algorithm>
#include <iomanip> //fixed<<setprecision(12)<<
#include <utility> //pair模板
#include <cmath>
#include <random> //mt19937 mtrand(time(0));
#include <time.h>
#define close                         \
    std::ios::sync_with_stdio(false); \
    cin.tie(0);                       \
    cout.tie(0)
typedef long long ll;
typedef unsigned long long ull;
const int N = 2e5 + 2;
const ll mod = 200907;
using namespace std;

int n, a, b;

void solve()
{
    //	freopen("data.in.txt","r",stdin);
    //	freopen("data.out.txt","w",stdout);
    cin >> n >> a >> b;
    if (a <= n / 2)
    {
        if (b <= n / 2)
        {
            for (int i = 1; i < n; i++)
                cout << "UL";
            for (int i = 1; i < b; i++)
                cout << 'R';
            for (int i = 1; i < a; i++)
                cout << 'D';
        }
        else
        {
            for (int i = 1; i < n; i++)
                cout << "UR";
            for (int i = 1; i < a; i++)
                cout << 'D';
            for (int i = 1; i <= n - b; i++)
                cout << 'L';
        }
    }
    else
    {
        if (b <= n / 2)
        {
            for (int i = 1; i < n; i++)
                cout << "DL";
            for (int i = 1; i < b; i++)
                cout << 'R';
            for (int i = 1; i <= n - a; i++)
                cout << 'U';
        }
        else
        {
            for (int i = 1; i < n; i++)
                cout << "DR";
            for (int i = 1; i <= n - a; i++)
                cout << 'U';
            for (int i = 1; i <= n - b; i++)
                cout << 'L';
        }
    }
}

int main()
{
    close;
    int t = 1;
    // cin>>t;
    while (t--)
    {
        solve();
    }
    return 0;
}

C. Klee in Solitary Confinement

对区间+k 找众数最多的情况

由于数的总数不多，因此考虑dp它+或者不+的状态实时更新最大值

用map会t 于是对每个数+2e6变成数组

#include<bits/stdc++.h>
#define close std::ios::sync_with_stdio(false),cin.tie(0),cout.tie(0)
using namespace std;
typedef long long ll;
const ll MAXN =  1e7+7;
const ll mod = 1e9+7;
const ll inf = 0x3f3f3f3f;
int mp[MAXN],tmp[MAXN],mx[MAXN]; 
int a[MAXN];
void solve(){
	int n,m;
	cin>>n>>m;
	for(int i=1;i<=n;i++){
		cin>>a[i];
		a[i]+=2e6; 
		mp[a[i]]++;
		tmp[a[i]+m]++;
		tmp[a[i]]--;
		if(tmp[a[i]]<0)
		tmp[a[i]]=0;
		mx[a[i]]=max(tmp[a[i]],mx[a[i]]);
		mx[a[i]+m]=max(tmp[a[i]+m],mx[a[i]+m]);
	} 
	int ans=0;
	for(int i=1;i<=n;i++){
		ans=max(ans,mp[a[i]]+mx[a[i]]);
	}
	cout<<ans;
}
signed main(){
	close;
	solve();
}

H. Crystalfly

树dp

在对u点进行dp的时候有两种情况

第一种是所有v都删除留个最大的 dp值由v的所有节点推出

第二种是留一个但是这个节点的子节点贡献不保留再由下面推出然后再加上一个t=3的节点然后再算v的子节点

取最大值就行没写明白wa了好久

#include<bits/stdc++.h>
#define close std::ios::sync_with_stdio(false),cin.tie(0),cout.tie(0)
using namespace std;
typedef long long ll;
const ll MAXN =  4e5+7;
const ll mod = 1e9+7;
const ll inf = 0x3f3f3f3f;
#define int ll
int a[MAXN],b[MAXN];
vector<int> adj[MAXN];
int dp[MAXN],sum[MAXN];
void dfs(int u,int fa) {
	dp[u]=a[u];
	int tmp=0;
	int max_1=0;
	int max_2=0;
	for(auto v:adj[u]) {
		if(v==fa) continue;
		sum[u]+=a[v];
		dfs(v,u);
		tmp=max(tmp,a[v]);
		if(b[v]==3) {
			if(a[v]>a[max_1]) {
				max_2=max_1;
				max_1=v;
			} else if(a[v]>a[max_2]) {
				max_2=v;
			}
		}
		dp[u]+=dp[v];
		dp[u]-=a[v];
	}
	int temp=dp[u];
	for(auto &v:adj[u]){
		if(v==fa) continue;
		int k=a[v];
		for(auto &it:adj[v]){
			if(it==u) continue;
			k+=dp[it]-a[it];
		}
		if(v==max_1)
		dp[u]=max({temp+tmp,temp-dp[v]+a[v]+k+a[max_2],dp[u]});
		else dp[u]=max({temp+tmp,temp-dp[v]+a[v]+k+a[max_1],dp[u]});
	}
	dp[u]=max(dp[u],temp+tmp);
}
void solve() {
	int n;
	cin>>n;
	for(int i=1; i<=n; i++) cin>>a[i];
	for(int i=1; i<=n; i++) {
		cin>>b[i];
		if(b[i]==2) b[i]=1;
		adj[i].clear();
		dp[i]=0;
		sum[i]=0;
	}
	for(int i=1; i<n; i++) {
		int u,v;
		cin>>u>>v;
		adj[u].push_back(v);
		adj[v].push_back(u);
	}
	sum[0]=inf;
	dfs(1,0);
	cout<<dp[1]<<'\n';
}
signed main() {
	close;
	int t;
	cin>>t;
	while(t--)
		solve();
}

M. Windblume Festival

手玩一下发现结论至少有一个改变符号然后最多有n-1个改变符号特判n=1的情况即可

改变符号贪心

#include <iostream>
#include <vector>
#include <stack>
#include <map>
#include <set>
#include <queue>
#include <string>
#include <algorithm>
#include <iomanip> //fixed<<setprecision(12)<<
#include <utility> //pair模板
#include <cmath>
#include <random> //mt19937 mtrand(time(0));
#include <time.h>
#define close                         \
    std::ios::sync_with_stdio(false); \
    cin.tie(0);                       \
    cout.tie(0)
typedef long long ll;
typedef unsigned long long ull;
const int N = 2e6 + 7;
const ll mod = 200907;
using namespace std;

ll n, a[N], ans, sum;

void solve()
{
    //	freopen("data.in.txt","r",stdin);
    //	freopen("data.out.txt","w",stdout);
    cin >> n;
    ans = sum = 0;
    for (int i = 1; i <= n; i++)
    {
        cin >> a[i];
    }
    if (n == 1)
    {
        cout << a[1] << endl;
        return;
    }
    sort(a + 1, a + 1 + n);
    for (int i = 1; i <= n; i++)
    {
        if (i == 1)
            sum -= a[i];
        else if (i == n)
            sum += a[i];
        else
        {
            if (a[i] < 0)
                sum -= a[i];
            else
                sum += a[i];
        }
    }
    cout << sum << endl;
}

int main()
{
    close;
    int t = 1;
    cin >> t;
    while (t--)
    {
        solve();
    }
    return 0;
}

E. Paimon Segment Tree

历史和是一种线性的递推可以由上一个递推过来由此想到矩阵处理因此用线段树维护矩阵加val之后相当于乘矩阵把求历史和过程中要用的全用矩阵维护好先

有点卡常少取模！！尽量优化下矩阵乘法只做一半之类的

#include<bits/stdc++.h>
#define close std::ios::sync_with_stdio(false),cin.tie(0),cout.tie(0)
using namespace std;
typedef long long ll;
const ll MAXN =  5e4+7;
const ll mod = 1e9+7;
const ll inf = 0x3f3f3f3f;
const ll M = 4;
inline ll read() { //快读
	ll x=0,f=0;
	char ch=getchar();
	while(ch>'9'||ch<'0') {
		f|=(ch=='-');
		ch=getchar();
	}
	while(ch<='9'&&ch>='0') {
		x=(x<<1LL)+(x<<3LL)+
		  (ch^48);
		ch=getchar();
	}
	return f?-x:x;
}
struct matrix {
	ll x[M+1][M+1];
	matrix() {
		memset(x,0,sizeof(x));
	}
	ll* operator [] (int i) {
		return x[i];
	}
	matrix operator * (matrix b) const {
		matrix res;
		for(int i = 1; i <= M; i ++) {
			for(int j = 1; j <= M; j ++) {
				for(int k = i; k <= j; k ++) {
					res[i][j] += (( x[i][k] *b[k][j]) );
					res[i][j]%=mod;
				}
			}
		}
		return res;
	}
	matrix operator + ( matrix b) const {
		matrix res;
		for (int i = 1; i <=M; i ++) {
			for (int j = i; j <= M; j ++) {
				res[i][j] = (x[i][j] + b[i][j] );
				res[i][j]%=mod;
			}
		}
		return res;
	}
	bool operator != (matrix b) {
		for (int i = 1; i <= M; i ++) {
			for (int j = 1; j <=M; j ++) {
				if (x[i][j] != b[i][j])return true;
			}
		}
		return false;
	}
	void init() {
		for(int i = 1 ; i <=M ; i ++) {
			for(int j = i ; j <=M ; j ++) {
				x[i][j] = (i == j ? 1LL : 0LL);
			}
		}
	}
};
matrix mpow(matrix a,ll m) { //矩阵a的m次方
	matrix res;
	res.init();
	while(m>0) {
		if(m&1) res=res*a;
		a=a*a;
		m>>=1;
	}
	return res;
}
bool check(matrix a) {
	int flag=true;
	for(int i=1; i<=M; i++) {
		for(int j=1; j<=M; j++) {
			if(i!=j&&a.x[i][j]!=0) flag=false;
			else if(i==j&&a.x[i][j]!=1) flag=false;
		}
	}
	return flag;
}
struct Info {
	matrix sum;
};
Info operator +(const Info &a,const Info &b) {
	Info c;
	c.sum=a.sum+b.sum;
	return c;
}
matrix A;
ll a[MAXN];
struct node {
	matrix lazy;
	Info val;
} seg[MAXN<<2];
void up(int id) {
	seg[id].val=seg[id<<1].val+seg[id<<1|1].val;
}
void settag(int id,int l,int r,matrix tag) {
	matrix tmp;
	seg[id].val.sum=seg[id].val.sum*tag;
	seg[id].lazy=seg[id].lazy*tag;
}
void down(int id,int l,int r) {
	if(seg[id].lazy.x[3][4]==0) return;
	int mid=l+r>>1;
	settag(id<<1,l,mid,seg[id].lazy);
	settag(id<<1|1,mid+1,r,seg[id].lazy);
	seg[id].lazy.init();
}
void build(int id,int l,int r) {
	seg[id].lazy.init();
	if(l==r) {
		seg[id].val.sum[1][1]=1;
		seg[id].val.sum[1][2]=a[l];
		seg[id].val.sum[1][3]=a[l]*a[l]%mod;
		seg[id].val.sum[1][4]=a[l]*a[l]%mod;
		return;
	}
	int mid = l+r>>1;
	build(id<<1,l,mid);
	build(id<<1|1,mid+1,r);
	up(id);
}
void modify(int id,int l,int r,int ql,int qr,matrix val) {
	if (ql==l&&r==qr) {
		settag(id,l,r,val);
		return;
	}
	down(id,l,r);
	int mid =(l+r) >> 1;
	if (qr<=mid)
		modify(id <<1,l,mid,ql,qr,val);
	else if (ql>mid)
		modify(id<<1|1, mid+1,r,ql,qr,val);
	else {
		modify(id<<1,l,mid,ql,mid,val);
		modify(id<<1|1,mid+1,r,mid+1,qr,val);
	}
	up(id);
}
Info query(int id,int l ,int r,int ql,int qr) {
	if(ql==l&&r==qr) {
		return seg[id].val;
	}
	down(id,l,r);
	int mid=l+r>>1;
	if(qr<=mid) return query(id<<1,l,mid,ql,qr);
	else if(ql>mid) return query(id<<1|1,mid+1,r,ql,qr);
	else return query(id<<1,l,mid,ql,mid)+query(id<<1|1,mid+1,r,mid+1,qr);
}
vector<array<int,3> > P;
vector<array<int,4> >  Q[MAXN];//id l r ty
ll ans[MAXN];
void solve() {
	int n,m,q;
//	cin>>n>>m>>q;
	n=read();
	m=read();
	q=read();
	for(int i=1; i<=n; i++) a[i]=read();
	build(1,1,n);
	P.push_back({0,0,0});
	for(int i=1; i<=m; i++) {
		int l,r,x;
//		cin>>l>>r>>x;
		l=read();
		r=read();
		x=read();
		P.push_back({l,r,x});
	}
	for(int i=1; i<=q; i++) {
		int l,r,x,y;
	//	cin>>l>>r>>x>>y;
		l=read();
		r=read();
		x=read();
		y=read();
		if(x!=0)
			Q[x-1].push_back({i,l,r,-1});
		Q[y].push_back({i,l,r,1});
	}
	for(auto it:Q[0]) {
		int id=it[0],l=it[1],r=it[2],ty=it[3];
		matrix tmp = query(1,1,n,l,r).sum;
		ans[id]+=ty*tmp[1][4];
	}
	for(int i=1; i<=m; i++) {
		int l=P[i][0],r=P[i][1],x=P[i][2];
		matrix tmp;
		tmp[1][1]=1;
		tmp[1][2]=x;
		tmp[1][3]=1LL*x*x%mod;
		tmp[1][4]=1LL*x*x%mod;
		tmp[2][2]=1;
		tmp[2][3]=2LL*x;
		tmp[2][4]=2LL*x;
		tmp[3][3]=1;
		tmp[3][4]=1;
		tmp[4][4]=1;
		modify(1,1,n,l,r,tmp);
		tmp.init();
		tmp[3][4]=1;
		if(l>1) {
			modify(1,1,n,1,l-1,tmp);
		}
		if(r<n) {
			modify(1,1,n,r+1,n,tmp);
		}
		for(auto it:Q[i]) {
			int id=it[0],l=it[1],r=it[2],ty=it[3];
			matrix tmp = query(1,1,n,l,r).sum;
			ans[id]+=ty*tmp[1][4];
		}
	}
	for(int i=1; i<=q; i++) {
		printf("%lld\n",(ans[i]%mod+mod)%mod);
	//	cout<<(ans[i]%mod+mod)%mod<<"\n";
	}
}
signed main() {
//	close;
	solve();
}

J. Xingqiu’s Joke

找因子用的很暴力不会证明复杂度的办法

4.2020-2021 ACM-ICPC, Asia Nanjing Regional Contest (XXI Open Cup, Grand Prix of Nanjing)

赛时过题EKL 补题AFHM

E. Evil Coordinate

重新排列路径然后让路径不会经过某个点

顺时针方向绕着走好像就行了赛时不小心写挂了好久

#include<bits/stdc++.h>
#define close std::ios::sync_with_stdio(false),cin.tie(0),cout.tie(0)
using namespace std;
typedef long long ll;
const ll MAXN =  3e5+7;
const ll mod = 1e9+7;
const ll inf = 0x3f3f3f3f;
#define int ll
int fx[4][2]= {1,0,0,1,-1,0,0,-1}; //右 上 左 下
int f[4];
string str[4]= {"R","U","L","D"};
void solve() {
	int x,y;
	cin>>x>>y;
	string s;
	cin>>s;
	string ans;
	for(int i=0; i<4; i++) f[i]=0;
	for(auto ch :s) {
		if(ch=='R') f[0]++;
		else if(ch=='U') f[1]++;
		else if(ch=='L') f[2]++;
		else f[3]++;
	}
	int xx=f[0]-f[2];
	int yy=f[1]-f[3];
	if((xx==x&&yy==y)||(x==0&&y==0)) {
		cout<<"Impossible\n";
		return;
	}
	int px=0,py=0;
	int flag1=1,flag2=1,flag3=1,flag4=1;
	for(int i=0;i<4;i++){
		for(int j=1;j<=f[i];j++){
			px+=fx[i][0];
			py+=fx[i][1];
			if(px==x&&py==y) flag1=0;
		}
	}
	
	px=0;
	py=0;
	for(int i=0;i<4;i++){
		int p=(i+1)%4; 
		for(int j=1;j<=f[p];j++){
			px+=fx[p][0];
			py+=fx[p][1];
			if(px==x&&py==y) flag2=0;
		}
	}
	px=0;
	py=0;
	for(int i=0;i<4;i++){
		int p=(i+2)%4; 
		for(int j=1;j<=f[p];j++){
			px+=fx[p][0];
			py+=fx[p][1];
			if(px==x&&py==y) flag3=0;
		}
	}
    px=0;
	py=0;
	for(int i=0;i<4;i++){
		int p=(i+3)%4; 
		for(int j=1;j<=f[p];j++){
			px+=fx[p][0];
			py+=fx[p][1];
			if(px==x&&py==y) flag4=0;
		}
	}
    // cout<<flag1<<" "<<flag2<<" "<<flag3<<" "<<flag4<<'\n';
	if(flag1){
		for(int i=0;i<4;i++){
			for(int j=1;j<=f[i];j++){
				cout<<str[i]; 
			}
		}
	}
	else if(flag2){
		for(int i=0;i<4;i++){
			int p=(i+1)%4;
			for(int j=1;j<=f[p];j++){
				cout<<str[p]; 
			}
		}
	}
	else if(flag3){
		for(int i=0;i<4;i++){
			int p=(i+2)%4;
			for(int j=1;j<=f[p];j++){
				cout<<str[p]; 
			}
		}
	} 
	else if(flag4){
		for(int i=0;i<4;i++){
			int p=(i+3)%4;
			for(int j=1;j<=f[p];j++){
				cout<<str[p]; 
			}
		}
	}
	else{
		cout<<"Impossible";
	}
	cout<<'\n';
}
signed main() {
	int t;
	cin>>t;
	while(t--)
		solve();
}

K. K Co-prime Permutation

显然当i位置上是i+1时 gcd必为1 否则为i时 gcd为i 推一下就行

#include<bits/stdc++.h>
#define close std::ios::sync_with_stdio(false),cin.tie(0),cout.tie(0)
using namespace std;
typedef long long ll;
const ll MAXN =  3e5+7;
const ll mod = 1e9+7;
const ll inf = 0x3f3f3f3f;
void solve(){
	int n,k;cin>>n>>k;
	if(k==0){
		cout<<"-1";
	} 
	else{
		k--;
		for(int i=1;i<=k;i++){
			cout<<i+1<<" ";
		}
		cout<<"1 ";
		for(int i=k+2;i<=n;i++){
			cout<<i<<" ";
		}
	}
}
signed main(){
	close;
	solve();
}

L. Let’s Play Curling

没记错就是找两个蓝色石头中间最大红色数量

#include <iostream>
#include <vector>
#include <stack>
#include <map>
#include <set>
#include <queue>
#include <string>
#include <algorithm>
#include <iomanip> //fixed<<setprecision(12)<<
#include <utility> //pair模板
#include <cmath>
#include <random> //mt19937 mtrand(time(0));
#include <time.h>
#define close                         \
    std::ios::sync_with_stdio(false); \
    cin.tie(0);                       \
    cout.tie(0)
typedef long long ll;
typedef unsigned long long ull;
const int inf = 2e9;
const int N = 2e6 + 7;
const ll mod = 200907;
using namespace std;

int n, m;
vector<int> all, a, b;

void solve()
{
    //	freopen("data.in.txt","r",stdin);
    //	freopen("data.out.txt","w",stdout);
    cin >> n >> m;
    all.clear(), a.clear(), b.clear();
    for (int i = 1; i <= n; i++)
    {
        int tp;
        cin >> tp;
        // cout << tp << ' ';
        a.push_back(tp);
        all.push_back(tp);
    }
    // cout << endl;
    for (int i = 1; i <= m; i++)
    {
        int tp;
        cin >> tp;
        b.push_back(tp);
        all.push_back(tp);
    }
    sort(a.begin(), a.end());
    sort(b.begin(), b.end());
    sort(all.begin(), all.end());
    all.erase(unique(all.begin(), all.end()), all.end());
    // for (int i = 0; i < all.size(); i++)
    // cout << all[i] << ' ';
    // cout << endl;
    for (int i = 0; i < a.size(); i++)
    {
        int tp = a[i];
        a[i] = lower_bound(all.begin(), all.end(), tp) - all.begin() + 1;
        // cout << a[i] << endl;
    }
    // cout << endl;
    for (int i = 0; i < b.size(); i++)
        b[i] = lower_bound(all.begin(), all.end(), b[i]) - all.begin() + 1;
    // cout << endl;
    int mx = 0;
    for (int i = 0; i < b.size() - 1; i++)
    {
        int l = upper_bound(a.begin(), a.end(), b[i]) - a.begin() + 1;
        int r = lower_bound(a.begin(), a.end(), b[i + 1]) - a.begin() + 1;
        // cout << l << ' ' << r << endl;
        mx = max(mx, r - l);
    }
    int l = upper_bound(a.begin(), a.end(), b[m - 1]) - a.begin() + 1;
    int r = lower_bound(a.begin(), a.end(), inf) - a.begin() + 1;
    mx = max(mx, r - l);
    l = 1;
    r = lower_bound(a.begin(), a.end(), b[0]) - a.begin() + 1;
    mx = max(mx, r - l);
    if (mx == 0)
        cout << "Impossible" << endl;
    else
        cout << mx << endl;
}

int main()
{
    close;
    int t = 1;
    cin >> t;
    while (t--)
    {
        solve();
    }
    return 0;
}

A. Ah, It’s Yesterday Once More

额场上写差不多了但是有一个孤立点没找到寄了就是斜着走

F. Fireworks

先推出期望计算公式然后三分答案场上三分太丑挂了

#include<bits/stdc++.h>
#define close std::ios::sync_with_stdio(false),cin.tie(0),cout.tie(0)
using namespace std;
typedef long long ll;
const ll MAXN =  3e5+7;
const ll mod = 1e9+7;
const ll inf = 0x3f3f3f3f;
#define double long double
double f(int x,int n,int m,double p){
	return ((double)x*n+m)/(1.0-(pow(1-p,x)));
}
void solve(){
	int n,m;
	double p;
	cin>>n>>m>>p;
	p/=10000.0;
	int l=0,r=inf;
	while(l<r){
		int mid1=l+(r-l)/3;
		int mid2=r-(r-l)/3;
		if(f(mid1,n,m,p)<f(mid2,n,m,p)) r=mid2-1;
		else l=mid1+1;
	}
	printf("%.8Lf\n",f(l,n,m,p));
}
signed main(){
	int t;cin>>t;
	while(t--)
	solve();
}

H. Harmonious Rectangle

小范围打表

M. Monster Hunter

感觉类似树上背包枚举父亲贡献的节点与我与我邻居贡献的

#include<bits/stdc++.h>
#define close std::ios::sync_with_stdio(false),cin.tie(0),cout.tie(0)
using namespace std;
typedef long long ll;
const ll MAXN =  3e3+7;
const ll mod = 1e9+7;
const ll inf = 0x3f3f3f3f;
#define int ll
int dp[MAXN][MAXN][2],a[MAXN];// i为顶点的子树 已经用了j个 此时用/不用  减少的最大贡献
vector<int> adj[MAXN];
int n;
int ans=0;
int sum[MAXN];
int sz[MAXN];
void dfs_sum(int u,int fa) {
	sum[u]=a[u];
	sz[u]=1;
	for(auto &v:adj[u]) {
		sum[u]+=a[v];
		dfs_sum(v,u);
	}
	ans+=sum[u];
}
void dfs(int u,int fa) {
	dp[u][1][1]=sum[u];
	for(auto &v:adj[u]) {
		dfs(v,u);
		for(int i=sz[u]; i>=0; i--) { //枚举父节点删了几个 
			for(int j=sz[v];j>=0;j--){
				if(j>0)
				dp[u][j+i][0]=max(dp[u][j+i][0],dp[u][i][0]+max(dp[v][j][0],dp[v][j][1]+a[v]));
				else 
				dp[u][j+i][0]=max(dp[u][j+i][0],dp[u][i][0]+dp[v][j][0]);
				if(j>0&&i>0)
				dp[u][j+i][1]=max(dp[u][j+i][1],dp[u][i][1]+max(dp[v][j][0],dp[v][j][1]));
				else if(i>0)
				dp[u][j+i][1]=max(dp[u][j+i][1],dp[u][i][1]+dp[v][j][0]);
			}
		}
		sz[u]+=sz[v]; 
	}

//	cout<<"delet : u "<<u<<"\n";
//	for(int i=0;i<=n;i++) cout<<"i = "<<i<<" "<<dp[u][i][0]<<" "<<dp[u][i][1]<<"\n";
}
void solve() {
	cin>>n;
	for(int i=1; i<=n; i++) {
		adj[i].clear();
		sum[i]=0;
		for(int j=0; j<=n; j++) dp[i][j][0]=dp[i][j][1]=0;
	}
	ans=0;
	for(int i=2; i<=n; i++) {
		int u;
		cin>>u;
		adj[u].push_back(i);
	}
	for(int i=1; i<=n; i++) cin>>a[i];
	dfs_sum(1,0);
	dfs(1,0);
	for(int i=0; i<=n; i++) {
		cout<<ans-max(dp[1][i][0],dp[1][i][1])<<" ";
	}
	cout<<"\n";
}
signed main() {
	int t;
	cin>>t;
	while(t--)
		solve();
}

5.2022 ICPC Southeastern Europe Regional Contest

赛时过题 AFGHN

A. AppendAppendAppend

一个一个遍历然后指针二分找下一个的位置超了+1

#include <iostream>
#include <vector>
#include <stack>
#include <map>
#include <set>
#include <queue>
#include <string>
#include <algorithm>
#include <iomanip> //fixed<<setprecision(12)<<
#include <utility> //pair模板
#include <cmath>
#include <random> //mt19937 mtrand(time(0));
#include <time.h>
#define close                         \
    std::ios::sync_with_stdio(false); \
    cin.tie(0);                       \
    cout.tie(0)
typedef long long ll;
typedef unsigned long long ull;
const int N = 2e5 + 7;
const ll mod = 200907;
using namespace std;

string s, t;
vector<int> pos[40];

void solve()
{
    //	freopen("data.in.txt","r",stdin);
    //	freopen("data.out.txt","w",stdout);
    cin >> s >> t;
    for (int i = 0; i < s.length(); i++)
        pos[s[i] - 'a'].push_back(i);
    int ans = 0, now = 0;
    for (int i = 0; i < t.length(); i++)
    {
        int j = t[i] - 'a';
        int nx = lower_bound(pos[j].begin(), pos[j].end(), now) - pos[j].begin();
        // cout << i << ' ' << now << ' ' << t[i] << ' ' << nx << endl;
        if (nx < pos[j].size())
            now = pos[j][nx] + 1;
        else
            now = 0, i--, ans++;
    }
    cout << ans + 1;
}

int main()
{
    close;
    int t = 1;
    // cin>>t;
    while (t--)
    {
        solve();
    }
    return 0;
}

F. Fortune over Sportsmanship

模拟一下怎么交换正解应该是dp 不过可能差不多

#include <iostream>
#include <vector>
#include <stack>
#include <map>
#include <set>
#include <queue>
#include <string>
#include <algorithm>
#include <iomanip> //fixed<<setprecision(12)<<
#include <utility> //pair模板
#include <cmath>
#include <random> //mt19937 mtrand(time(0));
#include <time.h>
#define close                         \
    std::ios::sync_with_stdio(false); \
    cin.tie(0);                       \
    cout.tie(0)
typedef long long ll;
typedef unsigned long long ull;
const int N = 2e3 + 7;
const ll mod = 200907;
using namespace std;

int n;
ll a[N][N];
vector<pair<int, int>> v;

void solve()
{
    //	freopen("data.in.txt","r",stdin);
    //	freopen("data.out.txt","w",stdout);
    cin >> n;
    for (int i = 1; i <= n; i++)
        for (int j = 1; j <= n; j++)
            cin >> a[i][j];
    ll ans = 0;
    for (int i = n; i > 1; i--)
    {
        ll mx = 0, idx = 0;
        for (int j = 1; j < i; j++)
            if (mx <= a[j][i])
                mx = a[j][i], idx = j;
        ans += mx;
        // cout << idx << ' ' << i << endl;
        v.push_back({idx, i});
        for (int j = 1; j < idx; j++)
        {
            a[j][idx] = max(a[j][idx], a[j][i]);
        }
        for (int j = idx + 1; j <= n; j++)
        {
            a[idx][j] = max(a[idx][j], a[j][i]);
        }
        // for (int j = 1; j <= n; j++)
        // {
        //     for (int k = 1; k <= n; k++)
        //         cout << a[j][k] << ' ';
        //     cout << endl;
        // }
        // cout << endl;
    }
    cout << ans << endl;
    for (auto i : v)
    {
        cout << i.first << ' ' << i.second << endl;
    }
}

int main()
{
    close;
    int t = 1;
    // cin>>t;
    while (t--)
    {
        solve();
    }
    return 0;
}

G. Gears

设第一个齿轮大小为x 后面的均能用x表示然后二分一下x的大小看看区间内是否有不满足的根据实际情况return true或者false

#include <iostream>
#include <vector>
#include <stack>
#include <map>
#include <set>
#include <queue>
#include <string>
#include <algorithm>
#include <iomanip> //fixed<<setprecision(12)<<
#include <utility> //pair模板
#include <cmath>
#include <random> //mt19937 mtrand(time(0));
#include <time.h>
#define close                         \
	std::ios::sync_with_stdio(false); \
	cin.tie(0);                       \
	cout.tie(0)
typedef long long ll;
typedef unsigned long long ull;
#define int ll
const int N = 5e5 + 7;
const ll mod = 200907;
using namespace std;
int n;
int a[N];
int b[N];
int c[N];
int R[N];
int sum=0;
int mins=1e9+7;
int maxs=0;
bool check(int mid) {
	int tmp=0;
	for(int i=1; i<=n; i++) {
		if(i%2==1) {
			tmp+=mid+c[i];
			if(mid+c[i]<mins){
				return true;
			}
			else if(mid+c[i]>maxs){
				return false;
			}
		} else {
			tmp+=c[i]-mid;
			if(c[i]-mid<mins){
				return false;
			}
			else if(c[i]-mid>maxs){
				return true; 
			}
		}
		//	cout<<"# "<<tmp<<'\n';
	}
	if(tmp>=sum) return false;
	else return true;
}
void solve() {
	cin>>n;
	for(int i=1; i<=n; i++) cin>>a[i];
	for(int i=1; i<=n; i++) {
		cin>>R[i];
		sum+=R[i];
		maxs=max(maxs,R[i]);
		mins=min(mins,R[i]);
	}
	for(int i=2; i<=n; i++) {
		b[i]=a[i]-a[i-1];
	}
	for(int i=1; i<=n; i++) {
		if(i==1) c[i]=0;
		else {
			c[i]=b[i]-c[i-1];
		}
	}
	int l=1,r=1e9+7;
	while(l<=r) {
		int mid=l+r>>1;
		if(check(mid)) {
			l=mid+1;
		} else {
			r=mid-1;
		}
	}
//	cout<<l<<" "<<r<<"\n";
	for(int i=1; i<=n; i++) {
		if(i%2==1) {
			cout<<l+c[i];
		} else cout<<c[i]-l;
		cout<<" ";
	}
	//	freopen("data.in.txt","r",stdin);
	//	freopen("data.out.txt","w",stdout);
}

signed main() {
	close;
	int t = 1;
	// cin>>t;
	while (t--) {
		solve();
	}
	return 0;
}

H. Hanoi

汉诺塔 1可以不受限制

#include <iostream>
#include <vector>
#include <stack>
#include <map>
#include <set>
#include <queue>
#include <string>
#include <algorithm>
#include <iomanip> //fixed<<setprecision(12)<<
#include <utility> //pair模板
#include <cmath>
#include <random> //mt19937 mtrand(time(0));
#include <time.h>
#define close                         \
    std::ios::sync_with_stdio(false); \
    cin.tie(0);                       \
    cout.tie(0)
typedef long long ll;
typedef unsigned long long ull;
const int N = 2e5 + 7;
const ll mod = 200907;
using namespace std;

int n;
stack<int> st1, st2, st3;
vector<string> ans;

void add(int k)
{
    stack<int> temp;
    while (st3.size() && k > st3.top())
    {
        temp.push(st3.top());
        st3.pop();
        ans.push_back("3 1");
    }
    st3.push(k);
    ans.push_back("2 3");
    while (temp.size())
    {
        st3.push(temp.top());
        temp.pop();
        ans.push_back("1 3");
    }
}

void solve()
{
    //	freopen("data.in.txt","r",stdin);
    //	freopen("data.out.txt","w",stdout);
    cin >> n;
    for (int i = 1, tp; i <= n; i++)
    {
        cin >> tp;
        st1.push(tp);
    }
    for (int i = n; i > 0; i--)
    {
        int tp = st1.top();
        st1.pop();
        ans.push_back("1 2");
        add(tp);
    }
    cout << ans.size() << endl;
    for (auto i : ans)
        cout << i << endl;
}

int main()
{
    close;
    int t = 1;
    // cin>>t;
    while (t--)
    {
        solve();
    }
    return 0;
}

N. Nusret Gökçe

让每个相邻的值的差不超过k 使用搜索分别往左往右递减遇到大的就从那个地方开始搜遇到小的并且搜过就return 应该是O（n)

#include <iostream>
#include <vector>
#include <stack>
#include <map>
#include <set>
#include <queue>
#include <string>
#include <algorithm>
#include <iomanip> //fixed<<setprecision(12)<<
#include <utility> //pair模板
#include <cmath>
#include <random> //mt19937 mtrand(time(0));
#include <time.h>
#define close                         \
    std::ios::sync_with_stdio(false); \
    cin.tie(0);                       \
    cout.tie(0)
typedef long long ll;
typedef unsigned long long ull;
const int N = 2e5 + 7;
const ll mod = 200907;
using namespace std;
#define int ll
int a[N],f[N];
int n,m;
void dfs(int x){
    int l=x,r=x;
    f[x]=max(f[x],a[x]);
    int now=a[x]-m;
    for(int i=r+1;i<=n;i++){
        if(f[i]<=now&&a[i]<=now){
             f[i]=now;
             now-=m;
        }
        else if(f[i]==0){
            dfs(i);
            break;
        }
        else if(f[i]>now) break;
    }
    now=a[x]-m;
    for(int i=l-1;i>=1;i--){
        if(f[i]<=now&&a[i]<=now){
            f[i]=now;
            now-=m;
        }
        else if(f[i]==0){
            dfs(i);
            break;
        }
        else if(f[i]>now) break;
    }
}
void solve()
{
    cin>>n>>m;
    int maxp=0;
    for(int i=1;i<=n;i++){
        cin>>a[i];
        if(a[i]>a[maxp]) maxp=i;
    }
    f[maxp]=a[maxp];
    dfs(maxp);
    for(int i=1;i<=n;i++){
        cout<<f[i]<<" ";
    }
}

signed main()
{
    close;
    int t = 1;
    // cin>>t;
    while (t--)
    {
        solve();
    }
    return 0;
}

6.The 15th Chinese Northeast Collegiate Programming Contest

赛时过题CEIKM 补题ACDHJ

C. Vertex Deletion

树dp 队友推了很牛的式子然后dp

#include <iostream>
#include <vector>
#include <stack>
#include <map>
#include <set>
#include <queue>
#include <string>
#include <algorithm>
#include <iomanip> //fixed<<setprecision(12)<<
#include <utility> //pair模板
#include <cmath>
#include <random> //mt19937 mtrand(time(0));
#include <time.h>
#define close                         \
    std::ios::sync_with_stdio(false); \
    cin.tie(0);                       \
    cout.tie(0)
typedef long long ll;
#define int ll
typedef unsigned long long ull;
const double pi = 3.141592653589793;
const int N = 2e5 + 7;
const int MAXN = 2e5 + 7;
const ll mod = 998244353;
using namespace std;
vector<int> adj[MAXN];
int dp[MAXN][3],sz[MAXN];
void dfs(int u,int fa){
	dp[u][0]=1;
	dp[u][1]=1;
	dp[u][2]=1;
	for(auto &v:adj[u]){
		if(v==fa) continue;
		dfs(v,u);
		dp[u][2]*=dp[v][0];
		dp[u][2]%=mod;
	}
	for(auto &v:adj[u]){
		if(v==fa) continue;
		dp[u][0]*=(dp[v][0]+dp[v][1]);
		dp[u][0]%=mod;
	}
	for(auto &v:adj[u]){
		if(v==fa) continue;
		dp[u][1]*=(dp[v][0]+dp[v][1]+dp[v][2]);
		dp[u][1]%=mod;
	}
	dp[u][1]-=dp[u][2];
	dp[u][1]%=mod;
	dp[u][1]=(dp[u][1]+mod)%mod;
//	cout<<"# u : "<<u<<"\n";
//	cout<<dp[u][0]<<" "<<dp[u][1]<<' '<<dp[u][2]<<'\n';
}
void solve()
{
    //	freopen("data.in.txt","r",stdin);
    //	freopen("data.out.txt","w",stdout);
    int n;
    cin>>n;
    for(int i=1;i<=n;i++) adj[i].clear();
    for(int i=1;i<n;i++){
    	int u,v;cin>>u>>v;
    	adj[u].push_back(v);
    	adj[v].push_back(u);
	}
	dfs(1,0);
	cout<<(dp[1][0]+dp[1][1])%mod<<"\n";
}

signed main()
{
    close;
    int t = 1;
     cin>>t;
    while (t--)
    {
        solve();
    }
    return 0;
}

E. Easy Math Problem

观察了一下发现要变成6的倍数

#include <iostream>
#include <vector>
#include <stack>
#include <map>
#include <set>
#include <queue>
#include <string>
#include <algorithm>
#include <iomanip> //fixed<<setprecision(12)<<
#include <utility> //pair模板
#include <cmath>
#include <random> //mt19937 mtrand(time(0));
#include <time.h>
#define close                         \
    std::ios::sync_with_stdio(false); \
    cin.tie(0);                       \
    cout.tie(0)
typedef long long ll;
typedef unsigned long long ull;
const double pi = 3.141592653589793;
const int N = 2e5 + 7;
const ll mod = 200907;
using namespace std;

void solve()
{
    ll n;
    cin >> n;
    cout << n * 6ll << ' ' << 3 << '\n';
    cout << 1ll * n << " " << 2ll * n << " " << 3ll * n << '\n';
    //	freopen("data.in.txt","r",stdin);
    //	freopen("data.out.txt","w",stdout);
}

int main()
{
    close;
    int t = 1;
    cin >> t;
    while (t--)
    {
        solve();
    }
    return 0;
}

I. Takeaway

特判签到

#include <iostream>
#include <vector>
#include <stack>
#include <map>
#include <set>
#include <queue>
#include <string>
#include <algorithm>
#include <iomanip> //fixed<<setprecision(12)<<
#include <utility> //pair模板
#include <cmath>
#include <random> //mt19937 mtrand(time(0));
#include <time.h>
#define close                         \
    std::ios::sync_with_stdio(false); \
    cin.tie(0);                       \
    cout.tie(0)
typedef long long ll;
#define int ll
typedef unsigned long long ull;
const double pi = 3.141592653589793;
const int N = 2e5 + 7;
const ll mod = 200907;
using namespace std;
int f[8] = {0, 7, 27, 41, 49, 63, 78, 108};
void solve()
{
    //	freopen("data.in.txt","r",stdin);
    //	freopen("data.out.txt","w",stdout);
    int n;
    cin >> n;
    int sum = 0;
    for (int i = 1; i <= n; i++)
    {
        int a;
        cin >> a;
        sum += f[a];
    }
    if (sum >= 120)
    {
        cout << sum - 50 << "\n";
    }
    else if (sum >= 89)
    {
        cout << sum - 30 << '\n';
    }
    else if (sum >= 69)
    {
        cout << sum - 15 << "\n";
    }
    else
        cout << sum << "\n";
}

signed main()
{
    close;
    int t = 1;
    cin >> t;
    while (t--)
    {
        solve();
    }
    return 0;
}

K. City

#include <iostream>
#include <vector>
#include <stack>
#include <map>
#include <set>
#include <queue>
#include <string>
#include <algorithm>
#include <iomanip> //fixed<<setprecision(12)<<
#include <utility> //pair模板
#include <cmath>
#include <random> //mt19937 mtrand(time(0));
#include <time.h>
#define close                         \
    std::ios::sync_with_stdio(false); \
    cin.tie(0);                       \
    cout.tie(0)
typedef long long ll;
typedef unsigned long long ull;
const double pi = 3.141592653589793;
const int N = 2e5 + 7;
const ll mod = 200907;
using namespace std;
#define int long long
struct node
{
    int x, y;
    int w;
    friend bool operator<(const node &a, const node &b)
    {
        return a.w > b.w;
    }
} edg[N];

struct quest
{
    int id;
    int w;
    int res;
    friend bool operator<(const quest &a, const quest &b)
    {
        return a.w > b.w;
    }
} que[N];

bool cmp(quest a, quest b)
{
    return a.id < b.id;
}

int p[N], siz[N];
int find(int x)
{
    if (p[x] != x)
        p[x] = find(p[x]);
    return p[x];
}

void solve()
{
    int n, m, q;
    cin >> n >> m >> q;
    for (int i = 1; i <= m; i++)
    {
        cin >> edg[i].x >> edg[i].y >> edg[i].w;
    }
    for (int i = 1; i <= q; i++)
    {
        cin >> que[i].w;
        que[i].id = i;
    }
    sort(edg + 1, edg + 1 + m);
    sort(que + 1, que + 1 + q);
    for (int i = 1; i <= n; i++)
    {
        p[i] = i;
        siz[i] = 1;
    }

    // for (int i = 1; i <= n; i++)
    // {
    //     cout << edg[i].w << ' ';
    // }
    // cout << '\n';
    // for (int i = 1; i <= q; i++)
    // {
    //     cout << que[i].w << ' ';
    // }
    // cout << '\n';

    int now = 1, ans = 0;
    for (int i = 1; i <= m; i++)
    {
        if (edg[i].w >= que[now].w)
        {
            int a = edg[i].x;
            int b = edg[i].y;
            if (find(a) == find(b))
            {
                continue;
            }
            // ans -= siz[find(b)] * (siz[find(b)] - 1) / 2;
            // ans -= siz[find(a)] * (siz[find(a)] - 1) / 2;
            ans += siz[find(b)] * siz[find(a)];
            siz[find(b)] += siz[find(a)];
            p[find(a)] = find(b);
            // ans += siz[find(b)] * (siz[find(b)] - 1) / 2;
        }
        else
        {
            que[now].res = ans;
            i--;
            now++;
            if (now > q)
                break;
        }
        // cout << ans << ' ';
    }
    // cout << now << '\n';
    for (int i = now; i <= q; i++)
    {
        // cout << i << "cnm";
        que[i].res = ans;
    }
    // cout << '\n';
    sort(que + 1, que + 1 + q, cmp);
    for (int i = 1; i <= q; i++)
    {
        cout << que[i].res << '\n';
    }
    //	freopen("data.in.txt","r",stdin);
    //	freopen("data.out.txt","w",stdout);
}

signed main()
{
    close;
    int t = 1;
    cin >> t;
    while (t--)
    {
        solve();
    }
    return 0;
}

M. Master of Shuangpin

写脑残了

#include<stdio.h>
#include<string.h>
char s[2000];
int main() {
	while(gets(s)!=NULL) {
		int n=strlen(s);
		int flag=1;
		for(int i=0; i<n; i++) {
			if(flag==1) {
				flag=0;
				if(s[i+1]==' '||i==n-1){
					printf("%c%c",s[i],s[i]);
					continue;
				}
				else if(s[i+2]==' '||i+1==n-1){
					printf("%c%c",s[i],s[i+1]);
					i++;
					continue;
				}
				
				if(s[i+1]=='h'){
					if(s[i]=='c') {
						if(s[i+2]==' '||i+1==n-1) printf("c");
						printf("i");
						i++;
						continue;
					} else if(s[i]=='s') {
						if(s[i+2]==' '||i+1==n-1) printf("s");
						printf("u");
						i++;
							continue;
					} else if(s[i]=='z') {
						if(s[i+2]==' '||i+1==n-1) printf("z");
						printf("v");
						i++;	continue;
					}
				} else if(s[i+1]=='i') {
					if(s[i]=='a') {
						if(s[i+2]==' '||i+1==n-1) printf("a");
						printf("d");
						i++;	continue;
					} else if(s[i]=='e') {
						if(s[i+2]==' '||i+1==n-1) printf("e");
						printf("w");
						i++;	continue;
					}
				} else if(s[i+1]=='n') {
					if(s[i]=='a'&&s[i+2]!='g') {
						if(s[i+2]==' '||i+1==n-1) printf("a");
						printf("j");
						i++;	continue;
					} else if(s[i]=='a'&&s[i+2]=='g') {
						if(s[i+3]==' '||i+2==n-1) printf("a");
						printf("h");
						i+=2;	continue;
					} else if(s[i]=='e'&&s[i+2]!='g') {
						if(s[i+2]==' '||i+1==n-1) printf("e");
						printf("f");
						i++;	continue;
					}
					else if(s[i]=='e'&&s[i+2]=='g'){
						if(s[i+3]==' '||i+2==n-1) printf("e");
							printf("g");
							i+=2;	continue;
						}
				} else if(s[i+1]=='u') {
					 if(s[i]=='o') {
					 	if(s[i+2]==' '||i+1==n-1) printf("o");
						printf("z");
						i++;	continue;
					}
				} 
				else if(s[i+1]=='o'){
					if(s[i]=='a'){
						if(s[i+2]==' '||i+1==n-1) printf("a");
						printf("c");
						i++;
						continue;
					}
				}
					printf("%c",s[i]);
			} else {
				if(s[i]==' ') {
					printf(" ");
					flag=1;
				} else {
					if(s[i+1]=='h') {
						if(s[i]=='c') {
							printf("i");
							i++;	continue;
						} else if(s[i]=='s') {
							printf("u");
							i++;	continue;
						} else if(s[i]=='z') {
							printf("v");
							i++;	continue;
						}
					} else if(s[i+1]=='i') {
						if(s[i]=='a') {
							printf("d");
							i++;	continue;
						} else if(s[i]=='e') {
							printf("w");
							i++;	continue;
						}
						else if(s[i]=='u'){
							printf("v");
							i++;
							continue;
						}
					} else if(s[i+1]=='n') {
						if(s[i]=='a'&&s[i+2]!='g') {
							printf("j");
							i++;	continue;
						} else if(s[i]=='a'&&s[i+2]=='g') {
							printf("h");
							i+=2;	continue;
						} else if(s[i]=='e'&&s[i+2]!='g') {
							printf("f");
							i++;	continue;
						}
						else if(s[i]=='e'&&s[i+2]=='g'){
							printf("g");
							i+=2;	continue;
						}
						else if(s[i]=='i'){
							if(s[i+2]=='g'){
								printf("k");
								i+=2;	continue;
							}
							else {
								printf("b");
								i+=1;	continue;
							}
						}
						else if(s[i]=='o'){
							if(s[i+2]=='g'){
								i+=2;	
								printf("s");continue;
							}
						}
						else if(s[i]=='u'){
							printf("y");
							i++;
							continue;
						}
					} else if(s[i+1]=='u') {
						if(s[i]=='i') {
							printf("q");
							i++;	continue;
						} else if(s[i]=='o') {
							printf("z");
							i++;	continue;
						}
					}
					else if(s[i+1]=='a'){
						if(s[i]=='u'){
							if(s[i+2]=='n'){
								if(s[i+3]=='g'){
									printf("l");
									i+=3;	continue;
								}
								else{
									printf("r");
									i+=2;	continue;
								}
							}
							else if(s[i+2]=='i'){
								printf("k");
								i+=2;	continue;
							}
							else{
								printf("x");
								i++;	continue;
							}
						}
						else if(s[i]=='i'){
							if(s[i+2]=='n'){
								if(s[i+3]=='g'){
									printf("l");
									i+=3;	continue;
								}
								else{
									printf("m");
									i+=2;	continue;
								}
							}
							else if(s[i+2]=='o'){
								printf("n");
								i+=2;	continue;
							}
							else {
								printf("x");
								i++;	continue;
							}
						}
					}
					else if(s[i+1]=='e'){
						if(s[i]=='i'){
							printf("p");
							i++;	continue;
						}
						else if(s[i]=='u'){
							printf("t");
							i++;	continue;
						}
					}
					else if(s[i+1]=='o'){
						if(s[i]=='u'){
							printf("o");
							i++;	continue;
						}
						else if(s[i]=='a'){
							printf("c");
							i++;	continue;
						}
						else if(s[i]=='i'){
							if(s[i+2]=='n'&&s[i+3]=='g'){
								printf("s");
								i+=3;	continue;
							}
						}
					}  
                        printf("%c",s[i]);
				}
			}
		}
		printf("\n");
	}
}

A. Matrix

组合数好像有优化先打完表然后直接输出也行

D. Lowbit

势能线段树

#include<bits/stdc++.h>
#define close std::ios::sync_with_stdio(false),cin.tie(0),cout.tie(0)
using namespace std;
typedef long long ll;
const ll MAXN =  3e5+7;
const ll mod = 998244353;
const ll inf = 0x3f3f3f3f;

#define int ll
int a[MAXN];
int lowbit(int x) {
	return x&-x;
}
int bitcnt(int x) {
	int cnt=0;
	while(x) {
		x-=lowbit(x);
		cnt++;
	}
	return cnt;
}
struct Info {
	int sum,cnt;
};
Info operator +(const Info &a,const Info &b) {
	Info c;
	c.sum=a.sum+b.sum;
	c.sum%=mod;
	c.cnt=a.cnt+b.cnt;
	return c;
}
struct node {
	int lazy;
	Info val;
} seg[MAXN<<2];
void up(int id) {
	seg[id].val=seg[id<<1].val+seg[id<<1|1].val;
}
void settag(int id,int l,int r,int val) {
	seg[id].val.sum*=val;
	seg[id].val.sum%=mod;
	seg[id].lazy*=val;
	seg[id].lazy%=mod;
}
void down(int id,int l,int r) {
	if(seg[id].lazy==1) return;
	int mid=l+r>>1;
	settag(id<<1,l,mid,seg[id].lazy);
	settag(id<<1|1,mid+1,r,seg[id].lazy);
	seg[id].lazy=1;
}
void build(int id,int l,int r) {
	seg[id].lazy=1;
	if(l==r) {
		seg[id].val.sum=a[l];
		seg[id].val.cnt=bitcnt(a[l]);
		return;
	}
	int mid = l+r>>1;
	build(id<<1,l,mid);
	build(id<<1|1,mid+1,r);
	up(id);
}
void change(int id,int l,int r,int pos) {
	if(r<pos||l>pos) return;
	if(l==r&&l==pos) {
		int k=lowbit(seg[id].val.sum);
		if(seg[id].val.cnt==1)
			seg[id].val.sum*=2,seg[id].val.sum%=mod;
		else seg[id].val.sum+=k,seg[id].val.cnt=bitcnt(seg[id].val.sum);
		return;
	}
	int mid=l+r>>1;
	down(id,l,r);
	change(id<<1,l,mid,pos);
	change(id<<1|1,mid+1,r,pos);
	up(id);
}
void modify(int id,int l,int r,int ql,int qr) {
	if (ql<=l&&r<=qr) {
		if(seg[id].val.cnt==r-l+1)
		settag(id,l,r,2);
		else{
			for(int i=l;i<=r;i++) change(id,l,r,i);
		}
		return;
	}
	down(id,l,r);
	int mid =(l+r) >> 1;
	if (qr<=mid)
		modify(id <<1,l,mid,ql,qr);
	else if (ql>mid)
		modify(id<<1|1, mid+1,r,ql,qr);
	else {
		modify(id<<1,l,mid,ql,qr);
		modify(id<<1|1,mid+1,r,ql,qr);
	}
	up(id);
}

Info query(int id,int l ,int r,int ql,int qr) {
	if(ql<=l&&r<=qr) {
		return seg[id].val;
	}
	down(id,l,r);
	int mid=l+r>>1;
	if(qr<=mid) return query(id<<1,l,mid,ql,qr);
	else if(ql>mid) return query(id<<1|1,mid+1,r,ql,qr);
	else return query(id<<1,l,mid,ql,qr)+query(id<<1|1,mid+1,r,ql,qr);
}
void solve() {
	int n;
	cin>>n;
	for(int i=1; i<=n; i++) cin>>a[i];
	build(1,1,n);
//	cout<<"# \n";
//	for(int i=1;i<=n;i++) cout<<query(1,1,n,i,i).sum<<" "<<query(1,1,n,i,i).cnt<<"\n";
	int q;
	cin>>q;
	while(q--) {
		int op,l,r;
		cin>>op>>l>>r;
		if(op==1) {
			modify(1,1,n,l,r);
		} else {
			cout<<query(1,1,n,l,r).sum%mod<<"\n";
		}
	}
}
signed main() {
	close;
	int t;
	cin>>t;
	while(t--)
		solve();
}

H. Loneliness

J. Transform

7.2021 Jiangxi Provincial Collegiate Programming Contest

赛时过题ABHJKL补题FG

A. Mio visits ACGN Exhibition

从左上角到右下角有多少条路径的1和0满足要求

dp i j k 现在的位置ij，现在路径的1数量的方案数

滚动数组优化下空间

#include <iostream>
#include <vector>
#include <stack>
#include <map>
#include <set>
#include <queue>
#include <string>
#include <algorithm>
#include <iomanip> //fixed<<setprecision(12)<<
#include <utility> //pair模板
#include <cmath>
#include <random> //mt19937 mtrand(time(0));
#include <time.h>
#define close                         \
	std::ios::sync_with_stdio(false); \
	cin.tie(0);                       \
	cout.tie(0)
typedef long long ll;
typedef unsigned long long ull;
const double pi = 3.141592653589793;
const int N = 2e5 + 7;
const ll mod = 998244353;
using namespace std;
ll dp[2][502][1002];
int mp[502][502];
void solve() {
	//	freopen("data.in.txt","r",stdin);
	//	freopen("data.out.txt","w",stdout);
	int m, n, p, q;
	cin >> n >> m >> p >> q;
	for (int i = 1; i <= n; i++) {
		for (int j = 1; j <= m; j++)
			cin >> mp[i][j];
	}
	if (p + q > n + m) {
		cout << "0";
		return;
	}
	for (int i = 1; i <= n; i++) {
		for (int j = 1; j <= n; j++) {
			if (i == 1 && j == 1) {
				dp[i%2][j][mp[i][j]] = 1;
				continue;
			}
			if (mp[i][j]== 0)
				for (int k = 0; k < i + j; k++) {
					dp[i%2][j][k] = dp[(i - 1)%2][j][k] + dp[i%2][(j - 1)][k];
					dp[i%2][j][k]%=mod;
			//		cout<<dp[i%2][j][k]<<" ";
				}
			else
				for (int k = 0; k < i + j; k++) {
					if(k==0) {
						dp[i%2][j][k]=0;
						continue;
					}
					dp[i%2][j][k] = dp[(i - 1)%2][j][k - 1] + dp[i%2][(j - 1)][k - 1];
					dp[i%2][j][k]%=mod;
	//			cout<<dp[i%2][j][k]<<" ";
				}
		}
	//	cout<<'\n';
	}
	ll ans=0;
	for(int i=q; i<n+m-p; i++) {
		ans+=dp[n%2][m][i];
	//	cout<<dp[n%2][m][i]<<"\n";
		ans%=mod;
	}
	cout << ans<<"\n";
}

int main() {
	close;
	int t = 1;
	// cin>>t;
	while (t--) {
		solve();
	}
	return 0;
}

B. Continued Fraction

#include <iostream>
#include <vector>
#include <stack>
#include <map>
#include <set>
#include <queue>
#include <string>
#include <algorithm>
#include <iomanip> //fixed<<setprecision(12)<<
#include <utility> //pair模板
#include <cmath>
#include <random> //mt19937 mtrand(time(0));
#include <time.h>
#define close                         \
    std::ios::sync_with_stdio(false); \
    cin.tie(0);                       \
    cout.tie(0)
typedef long long ll;
typedef unsigned long long ull;
const double pi = 3.141592653589793;
const int N = 2e2 + 7;
const ll mod = 200907;
using namespace std;
// int fac[N];
// void init()
// {
//     for (int i = 1; i < N; i++)
//     {
//         fac[i] = fac[i - 1] + i * i;
//     }
// }
void solve()
{
    int x, y;
    vector<int> ans;
    cin >> x >> y;
    while (1)
    {
        // cout << x << " " << y << '\n';
        ans.push_back(x / y);
        x %= y;
        if (x == 1)
            break;
        swap(x, y);
    }
    ans.push_back(y);
    cout << ans.size() - 1 << ' ';
    // if (x!=0)
    //     ans.push_back(y);
    // cout << ans.size() - 1 << ' ';
    for (auto x : ans)
    {
        cout << x << " ";
    }
    cout << '\n';
    //	freopen("data.in.txt","r",stdin);
    //	freopen("data.out.txt","w",stdout);
}

int main()
{
    // init();
    close;
    int t = 1;
    cin >> t;
    while (t--)
    {
        solve();
    }
    return 0;
}

H. Hearthstone So Easy

#include <iostream>
#include <vector>
#include <stack>
#include <map>
#include <set>
#include <queue>
#include <string>
#include <algorithm>
#include <iomanip> //fixed<<setprecision(12)<<
#include <utility> //pair模板
#include <cmath>
#include <random> //mt19937 mtrand(time(0));
#include <time.h>
#define close                         \
    std::ios::sync_with_stdio(false); \
    cin.tie(0);                       \
    cout.tie(0)
typedef long long ll;
typedef unsigned long long ull;
const double pi = 3.141592653589793;
const int N = 2e2 + 7;
const ll mod = 200907;
using namespace std;
// int fac[N];
// void init()
// {
//     for (int i = 1; i < N; i++)
//     {
//         fac[i] = fac[i - 1] + i * i;
//     }
// }
void solve()
{
    int n, k, fir = 0;
    cin >> n >> k;
    // if(n==1)fir=1;
    if (n > 1 && n <= 1 + k)
    {
        fir = 1;
    }
    puts(fir > 0 ? "pllj" : "freesin");
    //	freopen("data.in.txt","r",stdin);
    //	freopen("data.out.txt","w",stdout);
}

int main()
{
    // init();
    close;
    int t = 1;
    cin >> t;
    while (t--)
    {
        solve();
    }
    return 0;
}

J. LRU

二分空间大小 check里面O(n)跑就行了很快

#include <iostream>
#include <vector>
#include <stack>
#include <map>
#include <set>
#include <queue>
#include <string>
#include <algorithm>
#include <iomanip> //fixed<<setprecision(12)<<
#include <utility> //pair模板
#include <cmath>
#include <random> //mt19937 mtrand(time(0));
#include <time.h>
#define int long long
#define close                         \
    std::ios::sync_with_stdio(false); \
    cin.tie(0);                       \
    cout.tie(0)
typedef long long ll;
typedef unsigned long long ull;
const double pi = 3.141592653589793;
const int N = 2e5+ 7;
const ll mod = 200907;
using namespace std;

int n, k, a[N];
queue<pair<int, int>> que;
vector<int> X;
int getX(int x)
{
    return lower_bound(X.begin(), X.end(), x) - X.begin();
}
int mp[N], f[N];
bool check(int x)
{
    for (int i = 1; i <= n; i++)
        f[i] = 0, mp[i] = 0;
    int cnt = 0;
    while (que.size())
        que.pop();
    int sz = 0;
    for (int i = 1; i <= n; i++)
    {
        if (f[a[i]] == 1)
        {
            mp[a[i]] = i;
            cnt++;
        }
        else
        {
            sz++;
            f[a[i]] = 1;
            mp[a[i]] = i;
        }
        que.push({i, a[i]});
        while (sz > x)
        {
            int pos = que.front().first;
            int v = que.front().second;
            que.pop();
            if (pos == mp[v])
            {
                sz--;
                f[v] = 0;
            }
        }
        if (cnt >= k)
            return true;
    }
    if (cnt >= k)
        return true;
    else
        return false;
}
int bs1(int l, int r)
{ // 左偏二分
    while (l < r)
    {
        int mid = l + r >> 1;
        // cout << l << " " << r << " " << mid << " ";

        if (check(mid))
        {
            r = mid;
            // cout << "true\n";
        }
        else
        {
            l = mid + 1;
            // cout << "false\n";
        }
    }
    return l;
}
void solve()
{
    cin >> n >> k;
    X.push_back(-1);
    for (int i = 1; i <= n; i++)
    {
        cin >> a[i];
        X.push_back(a[i]);
    }
    sort(X.begin(), X.end());
    X.erase(unique(X.begin(), X.end()), X.end());
    for (int i = 1; i <= n; i++)
        a[i] = getX(a[i]);
    int ans = bs1(1, n + 1);
    if (ans != n + 1)
        cout << ans << '\n';
    else
        cout << "cbddl\n";

    //	freopen("data.in.txt","r",stdin);
    //	freopen("data.out.txt","w",stdout);
}

signed main()
{
    close;
    int t = 1;
    // cin >> t;
    while (t--)
    {
        solve();
    }
    return 0;
}

K. Many Littles Make a Mickle

#include <iostream>
#include <vector>
#include <stack>
#include <map>
#include <set>
#include <queue>
#include <string>
#include <algorithm>
#include <iomanip> //fixed<<setprecision(12)<<
#include <utility> //pair模板
#include <cmath>
#include <random> //mt19937 mtrand(time(0));
#include <time.h>
#define close                         \
    std::ios::sync_with_stdio(false); \
    cin.tie(0);                       \
    cout.tie(0)
typedef long long ll;
typedef unsigned long long ull;
const double pi = 3.141592653589793;
const int N = 2e2 + 7;
const ll mod = 200907;
using namespace std;
int fac[N];
void init()
{
    for (int i = 1; i < N; i++)
    {
        fac[i] = fac[i - 1] + i * i;
    }
}
void solve()
{
    int k, m;
    cin >> k >> m;
    cout << fac[k] * m << '\n';
    //	freopen("data.in.txt","r",stdin);
    //	freopen("data.out.txt","w",stdout);
}

int main()
{
    init();
    close;
    int t = 1;
    cin >> t;
    while (t--)
    {
        solve();
    }
    return 0;
}

L. It Rains Again

差分数组找有多少个点覆盖了

#include <iostream>
#include <vector>
#include <stack>
#include <map>
#include <set>
#include <queue>
#include <string>
#include <algorithm>
#include <iomanip> //fixed<<setprecision(12)<<
#include <utility> //pair模板
#include <cmath>
#include <random> //mt19937 mtrand(time(0));
#include <time.h>
#define close                         \
    std::ios::sync_with_stdio(false); \
    cin.tie(0);                       \
    cout.tie(0)
typedef long long ll;
typedef unsigned long long ull;
const double pi = 3.141592653589793;
const int N = 2e5 + 7;
const ll mod = 200907;
using namespace std;
int suf[N];
int pre[N];
void solve()
{
    //	freopen("data.in.txt","r",stdin);
    //	freopen("data.out.txt","w",stdout);
    int n;
    cin >> n;
    for (int i = 1; i <= n; i++)
    {
        int a, b, c, d;
        cin >> a >> b >> c >> d;
        suf[a]++;
        suf[c]--;
    }
    for (int i = 1; i <= 100007; i++)
    {
        pre[i] = pre[i - 1] + suf[i];
    }
    int ans = 0;
    for (int i = 1; i <= 100000; i++)
    {
        if (pre[i] > 0)
        {
            ans++;
            //  cout << "$ " << i << "\n";
        }
    }
    cout << ans;
}

int main()
{
    close;
    int t = 1;
    // cin>>t;
    while (t--)
    {
        solve();
    }
    return 0;
}

F. Four Column Hanoi Tower

打表然后观察发现第一次+2 第二次+2

+4+4+4+8+8+8+8这样的最后要输出一个大数

用py写的

arr = []
ans =[1]
tmp=1
for i in range(1,600):
    tmp*=2
    for j in range(0,i+1):
        arr.append(tmp)

for i in range(1,100000):
    ans.append(ans[i-1]+arr[i-1])

t=int(input())
for i in range(0,t):
    k=int(input())
    print(ans[k-1])

G. Magic Number Group

找最多的连续的因数出现的次数就像众数

那就用莫队

#include<bits/stdc++.h>
#define close std::ios::sync_with_stdio(false),cin.tie(0),cout.tie(0)
using namespace std;
typedef long long ll;
const ll MAXN =  1e6+700;
const ll mod = 1e9+7;
const ll inf = 0x3f3f3f3f;
#define int ll
bool isPri[MAXN]={1,1};//质数0 
int prime[MAXN];	int cnt=0;
void getPrime(){

	for(int i=2;i<=MAXN-5;i++){
		if(isPri[i]==0)  prime[++cnt]=i;
		for(int j=1;j<=cnt;j++){
			if(i*prime[j]>MAXN-5) break;
			isPri[i*prime[j]]=1;
			if(i%prime[j]==0) break;
		}
	}
} 
struct node{
	int L,R,k; 
}q[MAXN];
int pos[MAXN];
int a[MAXN];
int res[MAXN];
int ans=0,mx;	int n,m;
int coun[MAXN],ct[MAXN];
vector<int> vec[MAXN];
void init(){
	for(int i=1;i<=n;i++){
		int tmp=a[i];
		for(int j=1;j<cnt;j++){
			if(j*j>a[i]) break;
			if(tmp%prime[j]==0)  vec[i].push_back(prime[j]);
			while(tmp%prime[j]==0) tmp/=prime[j];
			if(tmp==1) break; 
		}
		if(tmp!=1) vec[i].push_back(tmp);
	}
}
bool cmp(node a,node b){
	if(pos[a.L]!=pos[b.L]) return pos[a.L]<pos[b.L];
	if(pos[a.L]&1) return a.R>b.R;
	return a.R<b.R; 
}
void add(int x){
	for(auto &it:vec[x]){
		ct[coun[it]]--;
		coun[it]++;
		ct[coun[it]]++;
		if(coun[it]>ans) ans=coun[it];
	}
}
void del(int x){
	for(auto &it:vec[x]){
		ct[coun[it]]--;
		coun[it]--;
		ct[coun[it]]++;
	}
	if(ct[ans]==0&&ans!=0) ans--;
}
void solve(){
	ans=0;
    cin>>n>>m;
	int B=n/sqrt(m)+1;
	for(int i=1;i<=n;i++) cin>>a[i];
	init();
	for(int i=1;i<=m;i++){
		cin>>q[i].L>>q[i].R;
		q[i].k=i; 
	}
	for(int i=1;i<=n;i++){
		pos[i]=(i-1)/B+1;
	}
	sort(q+1,q+1+m,cmp);
	int L=1,R=0;
	for(int i=1;i<=m;i++){
		while(L>q[i].L) add(--L);
		while(R<q[i].R) add(++R);
		while(L<q[i].L) del(L++);
		while(R>q[i].R) del(R--);
		res[q[i].k]=ans;
	}
	while(L<=R) del(L++); 
	for(int i=1;i<=m;i++) cout<<res[i]<<"\n";
	for(int i=1;i<=n;i++) vec[i].clear();
}
signed main(){
	close;
	getPrime();
	int t;cin>>t;
	while(t--)
	solve();
}

8.2019-2020 ACM-ICPC Pacific Northwest Regional Contest (Div. 1)

赛时过题ACDEM 补题B

A Radio Prize

换根板子需要推出一个能换的式子

#include <iostream>
#include <vector>
#include <stack>
#include <map>
#include <set>
#include <queue>
#include <string>
#include <algorithm>
#include <iomanip> //fixed<<setprecision(12)<<
#include <utility> //pair模板
#include <cmath>
#include <random> //mt19937 mtrand(time(0));
#include <time.h>
#define close                         \
    std::ios::sync_with_stdio(false); \
    cin.tie(0);                       \
    cout.tie(0)
typedef long long ll;
typedef unsigned long long ull;
const double pi = 3.141592653589793;
const int N = 2e5 + 7;
const ll mod = 998244353;
using namespace std;
#define int ll
int dp[2][N]; // sum path lenth
// int dp1[N];
// sz path lenth
int f[2][N];
int sum[N], sz[N];
int a[N];
vector<pair<int, int>> adj[N];
void dfs_down(int u, int fa)
{
    sz[u] = 1;
    sum[u] = a[u];
    int faw = 1;
    for (auto &it : adj[u])
    {
        int v = it.first, w = it.second;
        if (v == fa)
        {
            continue;
        }
        dfs_down(v, u);
        sz[u] += sz[v];
        sum[u] += sum[v];
        dp[0][u] += dp[0][v] + sum[v] * w;
        dp[1][u] += dp[1][v] + sz[v] * w;
    }
}
void dfs_up(int u, int fa)
{
    for (auto &it : adj[u])
    {
        int v = it.first, w = it.second;
        if (v == fa)
            continue;
        f[0][v] = f[0][u] + (sum[1] - 2 * sum[v]) * w;
        f[1][v] = f[1][u] + (sz[1] - 2 * sz[v]) * w;
        dfs_up(v, u);
    }
}
void solve()
{
    //	freopen("data.in.txt","r",stdin);
    //	freopen("data.out.txt","w",stdout);
    int n;
    cin >> n;
    for (int i = 1; i <= n; i++)
        cin >> a[i];
    for (int i = 1; i < n; i++)
    {
        int u, v, w;
        cin >> u >> v >> w;
        adj[u].push_back({v, w});
        adj[v].push_back({u, w});
    }
    dfs_down(1, 0);
    f[1][1] = dp[1][1];
    f[0][1] = dp[0][1];
    // cout<<"# "<<dp[1][1]<<' '<<dp[0][1]<<"\n";
    dfs_up(1, 0);
    for (int i = 1; i <= n; i++)
    {
        cout << f[1][i] * a[i] + f[0][i] << "\n";
    }
}

signed main()
{
    close;
    int t = 1;
    // cin>>t;
    while (t--)
    {
        solve();
    }
    return 0;
}

C Coloring Contention

找最大颜色变换的次数那从末尾开始dfs 然后找最短路径就行了

#include <iostream>
#include <vector>
#include <stack>
#include <map>
#include <set>
#include <queue>
#include <string>
#include <algorithm>
#include <iomanip> //fixed<<setprecision(12)<<
#include <utility> //pair模板
#include <cmath>
#include <random> //mt19937 mtrand(time(0));
#include <time.h>
#define close                         \
    std::ios::sync_with_stdio(false); \
    cin.tie(0);                       \
    cout.tie(0)
typedef long long ll;
typedef unsigned long long ull;
const double pi = 3.141592653589793;
const int N = 2e5 + 7;
const ll mod = 998244353;
using namespace std;
vector<int> adj[N];
int d[N], vis[N];
queue<int> que;
void solve()
{
    //	freopen("data.in.txt","r",stdin);
    //	freopen("data.out.txt","w",stdout);
    int n, m;
    cin >> n >> m;
    for(int i=1;i<=n;i++) d[i]=1e9+7;
    for (int i = 1; i <= m; i++)
    {
        int u, v;
        cin >> u >> v;
        adj[u].push_back(v);
        adj[v].push_back(u);
    }
    d[n] = 0;
    vis[n] = 1;
    que.push(n);
    while (que.size())
    {
        int u = que.front();
        que.pop();
        for (auto v : adj[u])
        {
            if (d[v] > d[u] + 1)
            {
                d[v] = d[u] + 1;
            }
            if (!vis[v])
            {
                que.push(v);
                vis[v] = 1;
            }
        }
    }
    cout << d[1]-1;
}

signed main()
{
    close;
    int t = 1;
    // cin>>t;
    while (t--)
    {
        solve();
    }
    return 0;
}

D Dividing by Two

贪心不停整除小了就加

#include <iostream>
#include <vector>
#include <stack>
#include <map>
#include <set>
#include <queue>
#include <string>
#include <algorithm>
#include <iomanip> //fixed<<setprecision(12)<<
#include <utility> //pair模板
#include <cmath>
#include <random> //mt19937 mtrand(time(0));
#include <time.h>
#define close                         \
    std::ios::sync_with_stdio(false); \
    cin.tie(0);                       \
    cout.tie(0)
typedef long long ll;
typedef unsigned long long ull;
const double pi = 3.141592653589793;
const int N = 2e5 + 7;
const ll mod = 998244353;
using namespace std;

ll a, b;

void solve()
{
    //	freopen("data.in.txt","r",stdin);
    //	freopen("data.out.txt","w",stdout);
    cin >> a >> b;
    ll ans = 0;
    while (a > b)
    {
        if (a % 2 == 0)
            a /= 2;
        else
            a++;
        ans++;
    }
    ans += b - a;
    cout << ans << endl;
}

signed main()
{
    close;
    int t = 1;
    // cin>>t;
    while (t--)
    {
        solve();
    }
    return 0;
}

E Rainbow Strings

#include <iostream>
#include <vector>
#include <stack>
#include <map>
#include <set>
#include <queue>
#include <string>
#include <algorithm>
#include <iomanip> //fixed<<setprecision(12)<<
#include <utility> //pair模板
#include <cmath>
#include <random> //mt19937 mtrand(time(0));
#include <time.h>
#define close                         \
    std::ios::sync_with_stdio(false); \
    cin.tie(0);                       \
    cout.tie(0)
typedef long long ll;
typedef unsigned long long ull;
const double pi = 3.141592653589793;
const int N = 2e5 + 7;
const ll mod = 11092019;
using namespace std;

string s;
int cnt[30];

void solve()
{
    string s;
    cin >> s;
    for (auto x : s)
    {
        cnt[x - 'a']++;
    }
    ll ans = 1;
    for (int i = 0; i < 26; i++)
    {
        ans *= (cnt[i] + 1);
        ans %= mod;
    }
    cout << ans << '\n';
    //	freopen("data.in.txt","r",stdin);
    //	freopen("data.out.txt","w",stdout);
    cin >> s;
}

signed main()
{
    close;
    int t = 1;
    // cin>>t;
    while (t--)
    {
        solve();
    }
    return 0;
}

M Maze Connect

找菱形的个数和由.开始搜索的连通块数量



#include <iostream>
#include <vector>
#include <stack>
#include <map>
#include <set>
#include <queue>
#include <string>
#include <algorithm>
#include <iomanip> //fixed<<setprecision(12)<<
#include <utility> //pair模板
#include <cmath>
#include <random> //mt19937 mtrand(time(0));
#include <time.h>
#define close                         \
    std::ios::sync_with_stdio(false); \
    cin.tie(0);                       \
    cout.tie(0)
typedef long long ll;
typedef unsigned long long ull;
const double pi = 3.141592653589793;
const int N = 1e3 + 7;
const ll mod = 998244353;
using namespace std;

int row, col;
char mp[N][N];
int vis[N][N];

int dx[4] = {-1, 0, 0, 1};
int dy[4] = {0, -1, 1, 0};
int ex[4] = {-1, 1, -1, 1};
int ey[4] = {-1, -1, 1, 1};

void solve()
{
    //	freopen("data.in.txt","r",stdin);
    //	freopen("data.out.txt","w",stdout);
    cin >> row >> col;
    int ans = 0;
    for (int i = 1; i <= row; i++)
        for (int j = 1; j <= col; j++)
            cin >> mp[i][j];
    if (row == 1)
    {
        for (int i = 1; i <= col; i++)
            if (mp[1][i] != '.')
                ans++;
        cout << ans << endl;
        return;
    }
    if (col == 1)
    {
        for (int i = 1; i <= row; i++)
            if (mp[i][1] != '.')
                ans++;
        cout << ans << endl;
        return;
    }
    // if(row==1&&col==1&&mp[1][1]!='.')cout<<1<<'\n';
    // if (mp[1][2] == '\\')
    //     cout << 'k' << endl;
    for (int i = 1; i <= row; i++)
        for (int j = 1; j <= col; j++)
        {
            if (mp[i][j] == '.' && vis[i][j] == 0)
            {
                bool flag = true;
                queue<pair<int, int>> q;
                q.push({i, j});
                while (q.size())
                {
                    int x = q.front().first, y = q.front().second;
                    q.pop();
                    // if (i == 2 && j == 2)
                    //     cout << x << ' ' << y << endl;
                    if (x <= 1 || y <= 1 || x >= row || y >= col)
                    {
                        flag = false;
                        continue;
                    }
                    if (vis[x][y] == 1)
                        continue;
                    vis[x][y] = 1;
                    for (int i = 0; i < 4; i++)
                    {
                        int xx = x + dx[i];
                        int yy = y + dy[i];
                        if (mp[xx][yy] != '\\' && mp[xx][yy] != '/')
                            q.push({xx, yy});
                    }
                    if (mp[x - 1][y - 1] != '\\' && mp[x - 1][y - 1] != '/')
                        if (mp[x][y - 1] != '/')
                            q.push({x - 1, y - 1});
                    if (mp[x - 1][y + 1] != '\\' && mp[x - 1][y + 1] != '/')
                        if (mp[x][y + 1] != '\\')
                            q.push({x - 1, y + 1});
                    if (mp[x + 1][y - 1] != '\\' && mp[x + 1][y - 1] != '/')
                        if (mp[x][y - 1] != '\\')
                            q.push({x + 1, y - 1});
                    if (mp[x + 1][y + 1] != '\\' && mp[x + 1][y + 1] != '/')
                        if (mp[x][y + 1] != '/')
                            q.push({x + 1, y + 1});
                }
                if (flag)
                    ans++;
            }
        }

    for (int i = 1; i < row; i++)
        for (int j = 1; j < col; j++)
            if (mp[i][j] == '/' && mp[i + 1][j] == '\\' && mp[i][j + 1] == '\\' && mp[i + 1][j + 1] == '/')
                ans++;
    cout << ans << endl;
}

signed main()
{
    close;
    int t = 1;
    // cin>>t;
    while (t--)
    {
        solve();
    }
    return 0;
}

B Perfect Flush

单调栈板子题没看出来

9.The 13th Chinese Northeast Collegiate Programming Contest

赛时过题BCEGHJ 补题DF

B. Balanced Diet

每个糖果选的个数不能少于limi或者不选

S为总的贡献值 C为数量最大的种类的数量

求S/C最大

前缀和求出每种糖果在有i个时的值然后遍历i

记得break 复杂度为O(n)

#include <iostream>
#include <vector>
#include <stack>
#include <map>
#include <set>
#include <queue>
#include <string>
#include <algorithm>
#include <iomanip> //fixed<<setprecision(12)<<
#include <utility> //pair模板
#include <cmath>
#include <random> //mt19937 mtrand(time(0));
#include <time.h>
#define close                         \
    std::ios::sync_with_stdio(false); \
    cin.tie(0);                       \
    cout.tie(0)
typedef long long ll;
typedef unsigned long long ull;
#define int ll
const double pi = 3.141592653589793;
const int N = 2e5 + 7;
const ll mod = 998244353;
using namespace std;
int l[N];
int prel[N];
vector<int> pre[N];
struct node{
	int len;
	int id;
}Node[N];
bool bj(node a,node b){
	return a.len>b.len;
}
void solve()
{
	int n,m;cin>>n>>m;
	for(int i=1;i<=m;i++){
		cin>>l[i];
	}
	for(int i=1;i<=m;i++){
		Node[i].id=i;
		
	}
	for(int i=1;i<=n;i++){
		int x,ty;cin>>x>>ty;
		pre[ty].push_back(x);
		Node[ty].len++;
	}
	prel[0]=m;
	for(int i=1;i<=m;i++){
		sort(pre[i].begin(),pre[i].end(),greater<int>());
		for(int j=1;j<min(l[i],Node[i].len);j++){
			pre[i][j]=pre[i][j]+pre[i][j-1];
		}
	}
	for(int i=1;i<=m;i++){
		prel[Node[i].len+1]--;
	}
	for(int i=1;i<=n;i++) prel[i]=prel[i-1]+prel[i];
	sort(Node+1,Node+m+1,bj);
	double ans=0.0;
	int ansx,ansy;int fz=0;
	for(int c=1;c<=n;c++){
		for(int now=1;now<=prel[c];now++){
			if(c<l[Node[now].id]) continue;
			fz+=pre[Node[now].id][c-1];
		}
		if(fz==0) continue;
		if(fz*1.0/c>ans){
			ans=fz*1.0/c;
			ansx=fz;
			ansy=c;
		}
	}
	int k=__gcd(ansx,ansy);
	ansx/=k;
	ansy/=k;
	cout<<ansx<<"/"<<ansy<<"\n";
	for(int i=1;i<=n;i++) prel[i]=0;
	for(int i=1;i<=m;i++) {
		pre[i].clear();
		Node[i].len=0;
	}
}

signed main()
{
    close;
    int t = 1;
    cin>>t;
    while (t--)
    {
        solve();
    }
    return 0;
}

C. Line-line Intersection

求有交点的线段的个数做了非正解不太好精度要开-22

#include <iostream>
#include <vector>
#include <stack>
#include <map>
#include <set>
#include <queue>
#include <string>
#include <algorithm>
#include <iomanip> //fixed<<setprecision(12)<<
#include <utility> //pair模板
#include <cmath>
#include <random> //mt19937 mtrand(time(0));
#include <time.h>
#define close                         \
    std::ios::sync_with_stdio(false); \
    cin.tie(0);                       \
    cout.tie(0)
typedef long long ll;
typedef unsigned long long ull;
const double pi = 3.141592653589793;
const int N = 2e5 + 7;
const ll inf = 0x3f3f3f3f;
const ll mod = 998244353;
const double eps = 1e-22;
using namespace std;
#define int ll
#define double long double
struct node
{
    double k;
    double b;
} edge[N];
bool bj(node a, node b)
{
    if (fabs(a.k - b.k) <= eps)
    {
        return a.b < b.b;
    }
    else
        return a.k < b.k;
}
void solve()
{
    //	freopen("data.in.txt","r",stdin);
    //	freopen("data.out.txt","w",stdout);
    int n;
    cin >> n;
    for (int i = 1; i <= n; i++)
    {
        int ax, ay, bx, by;
        cin >> ax >> ay >> bx >> by;
        if (bx != ax)
            edge[i].k = (by - ay) * 1.0 / (bx - ax) ,edge[i].b = ay - edge[i].k * ax;
        else
            edge[i].k = inf * 1.0,edge[i].b=ax;
    }
    sort(edge + 1, edge + n + 1, bj);
    ll ans = n * (n - 1) / 2, sav = 1, tol = 1, tmp = 0;
    for (int i = 2; i <= n; i++)
    {
        if (fabs(edge[i].k - edge[i - 1].k) >= eps)
        {
            tmp += sav * (sav - 1) / 2;
            // cerr << tol << " " << tmp << '\n';
            ans -= tol * (tol - 1) / 2 - tmp;
            // cout << "ans:" << ans << '\n';
            tol = 1;
            sav = 1;
            tmp = 0;
        }
        else if (fabs(edge[i].k - edge[i - 1].k) < eps && fabs(edge[i].b - edge[i - 1].b) < eps)
        {
            // cout<<"had ";
            tol++;
            sav++;
        }
        else
        {
            // cout << "sav:" << sav << '\n';
            tmp += sav * (sav - 1) / 2;
            sav = 1;
            tol++;
        }
    }
    tmp += sav * (sav - 1) / 2;
    // cerr << tol << " " << sav << '\n';

    ans -= tol * (tol - 1) / 2 - tmp;
    cout << ans << "\n";
}

signed main()
{
    close;
    int t = 1;
    cin >> t;
    while (t--)
    {
        solve();
    }
    return 0;
}

E. Minimum Spanning Tree

把树的边看成点然后建图最小生成树

每个子节点里面一层都是完全图完全图里所有边都取一次然后最小的取n-1次就行了

#include <iostream>
#include <vector>
#include <stack>
#include <map>
#include <set>
#include <queue>
#include <string>
#include <algorithm>
#include <iomanip> //fixed<<setprecision(12)<<
#include <utility> //pair模板
#include <cmath>
#include <random> //mt19937 mtrand(time(0));
#include <time.h>
#define close                         \
    std::ios::sync_with_stdio(false); \
    cin.tie(0);                       \
    cout.tie(0)
typedef long long ll;
typedef unsigned long long ull;
const double pi = 3.141592653589793;
const int N = 2e5 + 7;
const ll mod = 998244353;
using namespace std;
#define int long long
vector<pair<int, int>> G[N];
int d[N];
void solve()
{
    int n;
    cin >> n;
    for (int i = 1; i <= n; i++)
    {
        G[i].clear();
        d[i] = 0;
    }
    for (int i = 1; i < n; i++)
    {
        int x, y, w;
        cin >> x >> y >> w;
        G[x].push_back({y, w});
        G[y].push_back({x, w});
        d[x]++;
        d[y]++;
    }
    int ans = 0;
    for (int i = 1; i <= n; i++)
    {
        int sum = 0, minn = 1000000000000000;
        if (d[i] == 1)
            continue;
        if (d[i] == 2)
        {
            for (auto it : G[i])
            {
                int x = it.first, y = it.second;
                sum += y;
                minn = min(minn, y);
            }
            ans += sum;
            continue;
        }
        for (auto it : G[i])
        {
            int x = it.first, y = it.second;
            sum += y;
            minn = min(minn, y);
        }
        sum += minn * (d[i] - 2);
        ans += sum;
    }
    cout << ans << '\n';
    //	freopen("data.in.txt","r",stdin);
    //	freopen("data.out.txt","w",stdout);
}

signed main()
{
    close;
    int t = 1;
    cin >> t;
    while (t--)
    {
        solve();
    }
    return 0;
}

G. Radar Scanner

二分写的找x和y让大家最小的区间

#include <iostream>
#include <vector>
#include <stack>
#include <map>
#include <set>
#include <queue>
#include <string>
#include <algorithm>
#include <iomanip> //fixed<<setprecision(12)<<
#include <utility> //pair模板
#include <cmath>
#include <random> //mt19937 mtrand(time(0));
#include <time.h>
#define close                         \
    std::ios::sync_with_stdio(false); \
    cin.tie(0);                       \
    cout.tie(0)
typedef long long ll;
typedef unsigned long long ull;
const double pi = 3.141592653589793;
const int N = 2e5 + 7;
const ll inf = 1e9;
const ll mod = 998244353;
using namespace std;

int n;
struct rec
{
    ll x1, y1, x2, y2;
} rc[N];

struct qujian
{
    ll l, r;
} qj[N];

int find(ll pos)
{
    // cout << pos << endl;
    // for (int i = 1; i <= n; i++)
    //     cout << qj[i].l << ' ' << qj[i].r << endl;
    int cntl = 0, cntr = 0;
    for (int i = 1; i <= n; i++)
    {
        if (qj[i].r < pos)
            cntl++;
        else if (pos < qj[i].l)
            cntr++;
    }
    // cout << cntl << ' ' << cntr << endl;

    if (cntl >= cntr)
        return -1;
    else
        return 1;
}

ll count()
{
    ll l = 0, r = inf + 10, mid;
    while (l <= r)
    {
        mid = l + r >> 1;
        int op = find(mid);
        // cout << op << endl;
        // if (op == 0)
        //     break;
        if (op == -1)
            r = mid - 1;
        else
            l = mid + 1;
        // cout << l << ' ' << r << endl;
    }

    // cout << mid << endl;
    // cout << l << endl;
    ll ans = 0, mn;
    for (int i = 1; i <= n; i++)
    {
        if (qj[i].r < r)
            ans += r - qj[i].r;
        else if (r < qj[i].l)
            ans += qj[i].l - r;
    }
    mn = ans;
    ans = 0;
    for (int i = 1; i <= n; i++)
    {
        if (qj[i].r < l)
            ans += l - qj[i].r;
        else if (l < qj[i].l)
            ans += qj[i].l - l;
    }
    mn = min(mn, ans);
    return mn;
}

void solve()
{
    //	freopen("data.in.txt","r",stdin);
    //	freopen("data.out.txt","w",stdout);
    cin >> n;
    for (int i = 1; i <= n; i++)
        cin >> rc[i].x1 >> rc[i].y1 >> rc[i].x2 >> rc[i].y2;
    for (int i = 1; i <= n; i++)
        qj[i].l = rc[i].x1, qj[i].r = rc[i].x2;
    ll ans = 0;
    ans += count();

    for (int i = 1; i <= n; i++)
        qj[i].l = rc[i].y1, qj[i].r = rc[i].y2;
    ans += count();
    cout << ans << endl;
}

signed main()
{
    close;
    int t = 1;
    cin >> t;
    while (t--)
    {
        solve();
    }
    return 0;
}

H. Skyscraper

建造大楼会有操作抬升楼

最小建造次数的算法是所有上升的长度的和

就是对每个点求它和上一个点的差大于0有贡献

抬升区间(L-R) 只会改变L点和R+1点的贡献

用树状数组维护新数组和贡献

#include<bits/stdc++.h>
#include <iostream>
#include <vector>
#include <stack>
#include <map>
#include <set>
#include <queue>
#include <string>
#include <algorithm>
#include <iomanip> //fixed<<setprecision(12)<<
#include <utility> //pair模板
#include <cmath>
#include <random> //mt19937 mtrand(time(0));
#include <time.h>
#define close                         \
    std::ios::sync_with_stdio(false); \
    cin.tie(0);                       \
    cout.tie(0)
typedef long long ll;
typedef unsigned long long ull;
const double pi = 3.141592653589793;
const int N = 4e5 + 7;
const ll mod = 998244353;
using namespace std;
#define int ll
int t[N],a[N],t2[N];
int lowbit(int x){
	return x&-x;
}
ll getsum(int x){
	ll sum=0;
	while(x){
		sum+=t[x];
		x-=lowbit(x);
	}
	return sum;
}
void addv(int x,int val){
	while(x<=100005){
		t[x]+=val;
		x+=lowbit(x);
	}
}
ll getsum2(int x){
	ll sum=0;
	while(x){
		sum+=t2[x];
		x-=lowbit(x);
	}
	return sum;
}
void addv2(int x,int val){
	while(x<=100005){
		t2[x]+=val;
		x+=lowbit(x);
	}
}
void solve()
{
    //	freopen("data.in.txt","r",stdin);
    //	freopen("data.out.txt","w",stdout);
    memset(t,0,sizeof(t));
    memset(t2,0,sizeof(t));
    int n,m;cin>>n>>m;
    for(int i=1;i<=n;i++){
    	cin>>a[i];
	}
	for(int i=1;i<=n;i++){
		if(a[i]-a[i-1]>0) addv(i,a[i]-a[i-1]);
	}
	while(m--){
		int op;cin>>op;
		if(op==1){
			int l,r,x;cin>>l>>r>>x;
			addv2(l,x);
			addv2(r+1,-x);
			if(a[l]+getsum2(l)-a[l-1]-getsum2(l-1)>0){
				int tmp = getsum(l)-getsum(l-1);
				addv(l,a[l]+getsum2(l)-a[l-1]-getsum2(l-1)-tmp);
			}
			if(a[r]+getsum2(r)-a[r+1]-getsum2(r+1)>0){
				int tmp=getsum(r+1)-getsum(r);
				addv(r+1,tmp*-1);
			}
			else{
				int tmp=getsum(r+1)-getsum(r);
				addv(r+1,tmp*-1);
				addv(r+1,a[r+1]+getsum2(r+1)-(a[r]+getsum2(r)));
			}
		}
		else {
			int l,r;cin>>l>>r;
			cout<<a[l]+getsum2(l)+getsum(r)-getsum(l)<<"\n";
		}
//		for(int i=1;i<=n;i++){
//			cout<<a[i]+getsum2(i)<<" ";
//		}
//		cout<<'\n';
	}
}

signed main()
{
   close;
    int t = 1;
    cin>>t;
    while (t--)
    {
        solve();
    }
    return 0;
}

J. Time Limit

签到找tle时间

#include <iostream>
#include <vector>
#include <stack>
#include <map>
#include <set>
#include <queue>
#include <string>
#include <algorithm>
#include <iomanip> //fixed<<setprecision(12)<<
#include <utility> //pair模板
#include <cmath>
#include <random> //mt19937 mtrand(time(0));
#include <time.h>
#define close                         \
    std::ios::sync_with_stdio(false); \
    cin.tie(0);                       \
    cout.tie(0)
typedef long long ll;
typedef unsigned long long ull;
const double pi = 3.141592653589793;
const int N = 2e5 + 7;
const ll mod = 998244353;
using namespace std;

void solve()
{
    int n;
    cin >> n;

    vector<int> a(n + 10);
    for (int i = 1; i <= n; i++)
    {
        cin >> a[i];
    }
    int ans = a[1] * 3;
    for (int i = 2; i <= n; i++)
    {
        ans = max(ans, a[i] + 1);
    }
    if (ans % 2)
        cout << ans + 1 << '\n';
    else
        cout << ans << '\n';
    //	freopen("data.in.txt","r",stdin);
    //	freopen("data.out.txt","w",stdout);
}

signed main()
{
    close;
    int t = 1;
    cin >> t;
    while (t--)
    {
        solve();
    }
    return 0;
}

D. Master of Data Structure

场上还在瞎写

2000次询问是个虚树题建虚树暴力跑时间完全够

#include<bits/stdc++.h>
#define close std::ios::sync_with_stdio(false),cin.tie(0),cout.tie(0)
using namespace std;
typedef long long ll;
const ll MAXN =  5e5+7;
const ll mod = 1e9+7;
const ll inf = 0x3f3f3f3f;
#define int ll 
vector<array<int,4> > Q;
vector<int> adj[MAXN],vg[MAXN];
int par[MAXN][20],dep[MAXN];
int f[MAXN];
int w[MAXN];//u点向上的权值
int a[MAXN];
int d[MAXN];
//par[u][i]代表点u的祖先中 深度为max(1,dep[u]-2^i)是谁
int in[MAXN],out[MAXN],tim=0;
void dfs(int u,int fa) {
	dep[u]=dep[fa]+1;
	par[u][0]=fa;
	for(int i=1; i<20; ++i) {
		par[u][i]=par[par[u][i-1]][i-1];
	}
	for(auto &v:adj[u]) {
		if(v==fa) continue;
		dfs(v,u);
	}
}
int getLCA(int u,int v) {
	if(dep[u]<dep[v]) swap(u,v);
	for(int i=19; i>=0; --i) {
		if(dep[par[u][i]]>=dep[v]) u=par[u][i];
	}
	if(u==v) return u;
	for(int i=19; i>=0; i--) {
		if(par[u][i]!=par[v][i]) {
			u=par[u][i];
			v=par[v][i];
		}
	}
	return par[u][0];
}
void init(int u,int fa) {
	in[u]=++tim;
	for(auto v:adj[u]) {
		if(v==fa) continue;
		init(v,u);
	}
	out[u]=tim;
}
int isp(int u, int v) {
	return in[u] <= in[v] && out[v] <= out[u];
}
bool cmp(int u, int v) {
	return in[u] < in[v];
}
void build(vector<int>&node) {
	sort(node.begin(), node.end(), cmp);
	set<int>node_st;
	for (int x : node)node_st.insert(x);
	for (int i = 1; i < node.size(); i++)node_st.insert(getLCA(node[i - 1], node[i]));
	node.clear();
	for (int x : node_st)node.push_back(x);
	sort(node.begin(), node.end(), cmp);
	vector<int> st;
	for (int v : node) {
		while (!st.empty() && !isp(st.back(), v))
			st.pop_back();
		if (!st.empty()) {
			vg[st.back()].push_back(v);
		}

		st.push_back(v);
	}
}
void dfs2(int u,int fa) {
	d[u]=d[fa]+1;
	f[u]=fa;
	for(auto &v:vg[u]) {
		if(v==fa) continue;
		dfs2(v,u);
	}
}
//node.push_back(1)
//build(node);
void work(int u,int v,int k,int op) {
	int ans=0;
	if(op==1) {
		while(u!=v) {
			assert(u&&v);
			if(d[u]>d[v]) {
				a[u]+=k;
				w[u]+=k;
				u=f[u];
			} else {
				a[v]+=k;
				w[v]+=k;
				v=f[v];
			}
		}
		a[u]+=k;
	} else if(op==2) {
		while(u!=v) {
			assert(u&&v);
			if(d[u]>d[v]) {
				a[u]^=k;
				w[u]^=k;
				u=f[u];
			} else {
				a[v]^=k;
				w[v]^=k;
				v=f[v];
			}
		}
		a[u]^=k;
	} else if(op==3) {
		while(u!=v) {
			assert(u&&v);
			if(d[u]>d[v]) {
				if(a[u]>=k)
					a[u]-=k;
				if(w[u]>=k)
					w[u]-=k;
				u=f[u];
			} else {
				if(a[v]>=k)
					a[v]-=k;
				if(w[v]>=k)
					w[v]-=k;
				v=f[v];
			}
		}
		if(a[u]>=k) a[u]-=k;
	} else if(op==4) {
		while(u!=v) {
			assert(u&&v);
			if(d[u]>d[v]) {
				ans+=(dep[u]-dep[f[u]]-1)*w[u]+a[u];
				u=f[u];
			} else {
				ans+=(dep[v]-dep[f[v]]-1)*w[v]+a[v];
				v=f[v];
			}
		}
		ans+=a[u];
		cout<<ans<<"\n";
	} else if(op==5) {
		while(u!=v) {
			assert(u&&v);
			if(d[u]>d[v]) {
				ans^=a[u];
				if((dep[u]-dep[f[u]]-1)%2==1) ans^=w[u];
				u=f[u];
			} else {
				ans^=a[v];
				if((dep[v]-dep[f[v]]-1)%2==1) ans^=w[v];
				v=f[v];
			}
		}
		ans^=a[u];
		cout<<ans<<"\n";

	} else if(op==6) {
		int mins=a[u],maxs=a[u];
		while(u!=v) {
			assert(u&&v);
			if(d[u]>d[v]) {
				mins=min(mins,a[u]);
				if((dep[u]-dep[f[u]]-1))
					mins=min(mins,w[u]);
				maxs=max(maxs,a[u]);
				if((dep[u]-dep[f[u]]-1))
					maxs=max(maxs,w[u]);
				u=f[u];
			} else {
				mins=min(mins,a[v]);
				if((dep[v]-dep[f[v]]-1))
					mins=min(mins,w[v]);
				maxs=max(maxs,a[v]);
				if((dep[v]-dep[f[v]]-1))
					maxs=max(maxs,w[v]);
				v=f[v];
			}
		}
		mins=min(mins,a[u]);
		maxs=max(maxs,a[u]);
		ans=maxs-mins;
		cout<<ans<<'\n';

	} else if(op==7) {
		ans=inf;
		while(u!=v) {
			assert(u&&v);
			if(d[u]>d[v]) {
				ans=min(ans,abs(a[u]-k));
				if((dep[u]-dep[f[u]]-1))
					ans=min(ans,abs(w[u]-k));
				u=f[u];
			} else {
				ans=min(ans,abs(a[v]-k));
				if((dep[v]-dep[f[v]]-1))
					ans=min(ans,abs(w[v]-k));
				v=f[v];
			}
		}
		ans=min(ans,abs(a[u]-k));
		cout<<ans<<"\n";
	}
}
void solve() {
	int n,q;
	cin>>n>>q;
	vector<int> node;
	Q.clear();
	tim=0;
	for(int i=1; i<=n; i++) {
		for(int j=0; j<20; j++) par[i][j]=0;
		adj[i].clear();
		vg[i].clear();
		a[i]=0;
		w[i]=0;
		dep[i]=0;
		d[i]=0;
	}
	for(int i=1; i<n; i++) {
		int u,v;
		cin>>u>>v;
		adj[u].push_back(v);
		adj[v].push_back(u);
	}
	dfs(1,0);
	init(1,0);
	set<int> sta;
	for(int i=1; i<=q; i++) {
		int op;
		cin>>op;
		if(op==1||op==2||op==3||op==7) {
			int u,v,k;
			cin>>u>>v>>k;
			Q.push_back({op,u,v,k});
			sta.insert(u);
			sta.insert(v);
		} else {
			int u,v;
			cin>>u>>v;
			Q.push_back({op,u,v,0});
			sta.insert(u);
			sta.insert(v);
		}
	}
	sta.insert(1);
	assert(sta.size() <= 6000);
	for(auto &it:sta) node.push_back(it);
	build(node);
	dfs2(1,0);
	for(int i=0; i<q; i++) {
		int op=Q[i][0],u=Q[i][1],v=Q[i][2],k=Q[i][3];
		work(u,v,k,op);
//		cout<<"TEST : \n";
//		for(int i=1; i<=n; i++) {
//			cout<<"a: "<<a[i]<<" w: "<<w[i]<<"\n";
//		}
	}
}
signed main() {
    close;
	int t;
	cin>>t;
	while(t--)
		solve();
}

F. Mini-game Before Contest

图上博弈写SG函数一样的操作

10.The 10th Shandong Provincial Collegiate Programming Contest

赛时过题ACDFLM 补题EHIK

A. Calandar

给出日历规定算新日历

#include <iostream>
#include <vector>
#include <stack>
#include <map>
#include <set>
#include <queue>
#include <string>
#include <cstring>
#include <algorithm>
#include <iomanip> //fixed<<setprecision(12)<<
#include <utility> //pair模板
#include <cmath>
#include <random> //mt19937 mtrand(time(0));
#include <time.h>
#define close                         \
    std::ios::sync_with_stdio(false); \
    cin.tie(0);                       \
    cout.tie(0)
typedef long long ll;
typedef unsigned long long ull;
const double pi = 3.141592653589793;
const int inf = 2e9;
const int N = 2e5 + 7;
const ll mod = 998244353;
using namespace std;

ll y, m, d, cnt;
string a[5] = {"Monday", "Tuesday", "Wednesday", "Thursday", "Friday"};
string s;

void solve()
{
    //	freopen("data.in.txt","r",stdin);
    //	freopen("data.out.txt","w",stdout);
    cin >> y >> m >> d;
    cnt = d;
    cin >> s >> y >> m >> d;
    cnt = d - cnt;
    cnt %= 5;
    cnt = (cnt + 5) % 5;
    // cout << cnt << endl;
    int pos = 0;
    for (int i = 0; i < 5; i++)
        if (a[i] == s)
        {
            pos = i;
            break;
        }
    pos += cnt;
    pos %= 5;
    cout << a[pos] << endl;
}

signed main()
{
    close;
    int t = 1;
    cin >> t;
    while (t--)
    {
        solve();
    }
    return 0;
}

C. Wandering Robot

#include <iostream>
#include <vector>
#include <stack>
#include <map>
#include <set>
#include <queue>
#include <string>
#include <cstring>
#include <algorithm>
#include <iomanip> //fixed<<setprecision(12)<<
#include <utility> //pair模板
#include <cmath>
#include <random> //mt19937 mtrand(time(0));
#include <time.h>
#define close                         \
    std::ios::sync_with_stdio(false); \
    cin.tie(0);                       \
    cout.tie(0)
typedef long long ll;
typedef unsigned long long ull;
const double pi = 3.141592653589793;
const int inf = 2e9;
const int N = 2e5 + 7;
const ll mod = 998244353;
using namespace std;

ll len, n;
string s;

void solve()
{
    //	freopen("data.in.txt","r",stdin);
    //	freopen("data.out.txt","w",stdout);
    ll mx = 0;
    ll x = 0, y = 0;
    cin >> len >> n >> s;
    for (int i = 0; i < len; i++)
    {
        if (s[i] == 'U')
            y++;
        else if (s[i] == 'D')
            y--;
        else if (s[i] == 'L')
            x--;
        else if (s[i] == 'R')
            x++;
        mx = max(mx, abs(x) + abs(y));
    }
    x *= (n - 1);
    y *= (n - 1);
    for (int i = 0; i < len; i++)
    {
        if (s[i] == 'U')
            y++;
        else if (s[i] == 'D')
            y--;
        else if (s[i] == 'L')
            x--;
        else if (s[i] == 'R')
            x++;
        mx = max(mx, abs(x) + abs(y));
    }
    cout << mx << endl;
}

signed main()
{
    close;
    int t = 1;
    cin >> t;
    while (t--)
    {
        solve();
    }
    return 0;
}

D. Game on a Graph

取的最多个数就是m-n+1 看看谁取下一个另一个团队的人就赢了

#include <iostream>
#include <vector>
#include <stack>
#include <map>
#include <set>
#include <queue>
#include <string>
#include <cstring>
#include <algorithm>
#include <iomanip> //fixed<<setprecision(12)<<
#include <utility> //pair模板
#include <cmath>
#include <random> //mt19937 mtrand(time(0));
#include <time.h>
#define close                         \
    std::ios::sync_with_stdio(false); \
    cin.tie(0);                       \
    cout.tie(0)
typedef long long ll;
typedef unsigned long long ull;
const double pi = 3.141592653589793;
const int inf = 2e9;
const int N = 2e5 + 7;
const ll mod = 998244353;
using namespace std;

void solve()
{
    //	freopen("data.in.txt","r",stdin);
    //	freopen("data.out.txt","w",stdout);
    int k;
    cin>>k;
    string s;cin>>s;s=" "+s;
    int n,m;cin>>n>>m;
    for(int i=1;i<=m;i++){
    	int u,v;cin>>u>>v;
	}
	int tmp=m-n+1;
	tmp++;
	tmp=(tmp-1)%k+1;
	if(s[tmp]=='1') cout<<"2\n";
	else cout<<"1\n";
	
}

signed main()
{
    close;
    int t = 1;
    cin>>t;
    while (t--)
    {
        solve();
    }
    return 0;
}

F. Stones in the Bucket

让所有数字一样至少需要几步

贪心最少步数的时候肯定是每个数在到达可能最大的平均值

#include <iostream>
#include <vector>
#include <stack>
#include <map>
#include <set>
#include <queue>
#include <string>
#include <cstring>
#include <algorithm>
#include <iomanip> //fixed<<setprecision(12)<<
#include <utility> //pair模板
#include <cmath>
#include <random> //mt19937 mtrand(time(0));
#include <time.h>
#define close                         \
    std::ios::sync_with_stdio(false); \
    cin.tie(0);                       \
    cout.tie(0)
typedef long long ll;
typedef unsigned long long ull;
#define int ll
const double pi = 3.141592653589793;
const int inf = 2e9;
const int N = 2e5 + 7;
const ll mod = 998244353;
using namespace std;
int a[N];
void solve()
{
    //	freopen("data.in.txt","r",stdin);
    //	freopen("data.out.txt","w",stdout);
    int n;cin>>n;
    int sum=0;
    for(int i=1;i<=n;i++){
    	cin>>a[i];
    	sum+=a[i];
	}
	int ans=0;
	sum/=n;
	for(int i=1;i<=n;i++){
		if(a[i]>sum){
			ans+=a[i]-sum;
		}
	}
	cout<<ans<<'\n';
}

signed main()
{
    close;
    int t = 1;
     cin>>t;
    while (t--)
    {
        solve();
    }
    return 0;
}

L. Median

给定数大小关系看某个数是否可能变成中位数

用弗洛伊德算法求传递凸包，如果一个数到达它的数不超过n/2 它到达的也不超过n/2就可以

如果某个数既能到达它它又能到达那就是不可能的

（无视这个tarjan

#include <iostream>
#include <vector>
#include <stack>
#include <map>
#include <set>
#include <queue>
#include <string>
#include <cstring>
#include <algorithm>
#include <iomanip> //fixed<<setprecision(12)<<
#include <utility> //pair模板
#include <cmath>
#include <random> //mt19937 mtrand(time(0));
#include <time.h>
#define close                         \
    std::ios::sync_with_stdio(false); \
    cin.tie(0);                       \
    cout.tie(0)
typedef long long ll;
typedef unsigned long long ull;
const double pi = 3.141592653589793;
const int inf = 0x3f3f3f3f;
const int N = 2e5 + 7;
const ll mod = 998244353;
using namespace std;
int ans[N];
int d[105][105];
int cnt;
int low[N], num[N], dfn;
int sccno[N], sta[N], top;
vector<int> G[N];
void dfs(int u)
{
    sta[top++] = u;
    low[u] = num[u] = ++dfn;
    for (int i = 0; i < G[u].size(); i++)
    {
        int v = G[u][i];
        if (!num[v])
        {
            dfs(v);
            low[u] = min(low[v], low[u]);
        }
        else if (!sccno[v])
        {
            low[u] = min(low[u], num[v]);
        }
    }
    if (low[u] == num[u])
    {
        cnt++;
        while (1)
        {
            int v = sta[--top];
            sccno[v] = cnt;
            if (u == v)
                break;
        }
    }
}
void Tarjan(int n)
{
    cnt = top = dfn = 0;
    memset(sccno, 0, sizeof sccno);
    memset(num, 0, sizeof num);
    memset(low, 0, sizeof low);
    for (int i = 1; i <= n; i++)
    {
        if (!num[i])
        {
            dfs(i);
        }
    }
}

void solve()
{
    int n, m, vaild = 1;
    cin >> n >> m;
    memset(d, 0x3f3f3f3f, sizeof(d));
    for (int i = 0; i <= n; i++)
    {
        G[i].clear();
    }
    for (int i = 1; i <= m; i++)
    {
        int x, y;
        cin >> x >> y;
        if (x == y)
            vaild = 0;
        G[y].push_back(x); // taj
        d[x][y] = 1;
    }

    Tarjan(n);
    if (cnt != n)
        vaild = 0;

    if (vaild == 0)
    {
        for (int i = 1; i <= n; i++)
        {
            cout << '0';
        }
        cout << '\n';
        return;
    }
    for (int k = 1; k <= n; k++)
    {
        for (int u = 1; u <= n; u++)
        {
            for (int v = 1; v <= n; v++)
            {
                d[u][v] = min(d[u][v], d[u][k] + d[k][v]);
            }
        }
    }
    for (int i = 1; i <= n; i++)
    {
        int In = 0, Out = 0;
        for (int j = 1; j <= n; j++)
        {
            if (i == j)
                continue;
            if (d[i][j] != inf && d[j][i] != inf)
                vaild = 0;
            if (d[i][j] != inf)
                In++;
            if (d[j][i] != inf)
                Out++;
            if (In <= (n / 2) && Out <= (n / 2))
                ans[i] = 1;
            else
                ans[i] = 0;
        }
    }
    if (vaild == 0)
    {
        for (int i = 1; i <= n; i++)
        {
            cout << '0';
        }
        cout << '\n';
        return;
    }
    for (int i = 1; i <= n; i++)
        cout << ans[i];
    cout << "\n";
    //	freopen("data.in.txt","r",stdin);
    //	freopen("data.out.txt","w",stdout);
}

signed main()
{
    close;
    int t = 1;
    cin >> t;
    while (t--)
    {
        solve();
    }
    return 0;
}

M. Sekiro

钱的数量超出就直接输出1 其他的暴力跑复杂度不高

#include <iostream>
#include <vector>
#include <stack>
#include <map>
#include <set>
#include <queue>
#include <string>
#include <cstring>
#include <algorithm>
#include <iomanip> //fixed<<setprecision(12)<<
#include <utility> //pair模板
#include <cmath>
#include <random> //mt19937 mtrand(time(0));
#include <time.h>
#define close                         \
    std::ios::sync_with_stdio(false); \
    cin.tie(0);                       \
    cout.tie(0)
typedef long long ll;
typedef unsigned long long ull;
#define int long long
const double pi = 3.141592653589793;
const int inf = 2e9;
const int N = 2e5 + 7;
const ll mod = 998244353;
using namespace std;
int fac[40];
void init()
{
    fac[0] = 1;
    for (int i = 1; i <= 30; i++)
    {
        fac[i] = fac[i - 1] * 2;
    }
}
void solve()
{
    int n;
    int k;
    cin >> n >> k;
    if (n == 0)
    {
        cout << 0 << "\n";
        return;
    }
    if (k >= 31)
    {
        cout << 1 << '\n';
        return;
    }
    for (int i = 1; i <= k; i++)
    {
        if (n % 2)
            n = n / 2 + 1;
        else
            n /= 2;
    }

    cout << n << '\n';
    //	freopen("data.in.txt","r",stdin);
    //	freopen("data.out.txt","w",stdout);
}

signed main()
{
    close;
    init();
    int t = 1;
    cin >> t;
    while (t--)
    {
        solve();
    }
    return 0;
}

E. BaoBao Loves Reading

和之前那个好像但是不能那样想

这次要考虑某一个数有贡献的条件是容量大于中间不同数的数量

那么我们就算出每一个数什么时候有贡献就行了

经典树状数组扫描线求数种类

#include<bits/stdc++.h>
#define close std::ios::sync_with_stdio(false),cin.tie(0),cout.tie(0)
using namespace std;
typedef long long ll;
const ll MAXN =  3e5+7;
const ll mod = 1e9+7;
const ll inf = 0x3f3f3f3f;
int last[MAXN],ans[MAXN]; 
struct BIT {
	int t[MAXN]={0};
	BIT(){
		memset(t,0,sizeof(t));
	}
	int lowbit(int x) {
		return x&-x;
	}
	ll getsum(int x) {
		ll sum=0;
		while(x) {

			sum+=t[x];
			x-=lowbit(x);
		}
		return sum;
	}
	void addv(int x,int val) {
		while(x<=MAXN) {
			t[x]+=val;
			x+=lowbit(x);
		}
	}
};
void solve(){
	BIT T;
	int n;cin>>n;
	for(int i=1;i<=n;i++) last[i]=0,ans[i]=0;
	for(int i=1;i<=n;i++){
		int x;cin>>x;
		if(last[x]==0){
			T.addv(i,1);
		}
		else{
			T.addv(last[x],-1);
			T.addv(i,1);
			int sum=T.getsum(i)-T.getsum(last[x]-1);
			ans[sum]++;
		}
		last[x]=i;
	}
	for(int i=1;i<=n;i++) ans[i]=ans[i-1]+ans[i];
	for(int i=1;i<=n;i++) {
		cout<<n-ans[i];
		if(i!=n) cout<<" ";
	}
	cout<<'\n'; 
}
signed main(){
    close;
	int t;cin>>t;
	while(t--)
	solve();
}

H. Tokens on the Segments

每个线段放一个石头就会贡献+1，但是不能在一个位置放两个石头，问最多贡献

用set模拟也可以做

用优先队列方便先取最左边的可以取就丢掉不能取就把它的头-1

#include<bits/stdc++.h>
#define close std::ios::sync_with_stdio(false),cin.tie(0),cout.tie(0)
using namespace std;
typedef long long ll;
const ll MAXN =  3e5+7;
const ll mod = 1e9+7;
const ll inf = 0x3f3f3f3f;
#define int ll
priority_queue<pair<int,int> > que;
void solve(){
	int n;cin>>n; 
	while(que.size()) que.pop();
	for(int i=1;i<=n;i++){
		int l,r;cin>>l>>r;
		que.push({l*-1,r*-1});
	}
	int ans=0,now=0;
	while(que.size()){
		auto st=que.top();
		que.pop();
		if(st.first*-1>now){
			now=st.first*-1;
			ans++;
		}
		else {
			int x=st.first;
			x*=-1;
			x++;
			int y=st.second;
			y*=-1;
			if(x>y);
			else que.push({x*-1,y*-1});
		}
	}
	cout<<ans<<'\n';
}
signed main(){
    close;
	int t;cin>>t;
	while(t--)
	solve();
}

I. Connected Intervals

要统计出符合条件的区间的数量，意思是（L，R）是从L-R的所有点

只看这些点的时候如果连通块数量是1 就符合条件

考虑扫描线在加入一个数 3 的时候假设3是一个完全独立的点不与其他数相连所有区间如(1,3),(2,3),(3,3)的连通块数量都会+1

但是如果告诉你 3和2连接那3加入（2，3)的时候其实连通块数量没有+1 那就给2，3的数量-1 然后1，3的同理也要-1

设线段树上的点 1 2 3 4 ...... 表示区间的左端点扫到3 就表示到3位为止的区间，然后up的时候统计线段内最小值和最小值的数量

如果最小值==1 就return cnt，意思就是让线段树来找为1 的点的个数

#include<bits/stdc++.h>
#define close std::ios::sync_with_stdio(false),cin.tie(0),cout.tie(0)
using namespace std;
#define int ll
typedef long long ll;
const ll MAXN =  3e5+7;
const ll mod = 1e9+7;
const ll inf = 0x3f3f3f3f;
vector<int> adj[MAXN];
struct Info {
	int mn,cnt;
};
Info operator +(const Info &a,const Info &b) {
	Info c;
	c.mn=min(a.mn,b.mn);
	c.cnt=0;
	if(c.mn==a.mn) c.cnt+=a.cnt;
	if(c.mn==b.mn) c.cnt+=b.cnt;
	return c;
}
struct node {
	int lazy;
	Info val;
} seg[MAXN<<2];
void up(int id) {
	seg[id].val=seg[id<<1].val+seg[id<<1|1].val;
}
void settag(int id,int l,int r,int tag) {
	seg[id].val.mn+=tag;
	seg[id].lazy+=tag;
}
void down(int id,int l,int r) {
	if(seg[id].lazy==0) return;
	int mid=l+r>>1;
	settag(id<<1,l,mid,seg[id].lazy);
	settag(id<<1|1,mid+1,r,seg[id].lazy);
	seg[id].lazy=0;
}
void build(int id,int l,int r) {
	seg[id].lazy=0;
	seg[id].val.cnt=0;
	if(l==r) {
		seg[id].val.mn=0;
		seg[id].val.cnt=1;
		return;
	}
	int mid = l+r>>1;
	build(id<<1,l,mid);
	build(id<<1|1,mid+1,r);
	up(id);
}
void modify(int id,int l,int r,int ql,int qr,int val) {
	if (ql==l&&r==qr) {
		settag(id,l,r,val);
		return;
	}
	down(id,l,r);
	int mid =(l+r) >> 1;
	if (qr<=mid)
		modify(id <<1,l,mid,ql,qr,val);
	else if (ql>mid)
		modify(id<<1|1, mid+1,r,ql,qr,val);
	else {
		modify(id<<1,l,mid,ql,mid,val);
		modify(id<<1|1,mid+1,r,mid+1,qr,val);
	}
	up(id);
}
int query(int id,int l ,int r,int ql,int qr) {
	if(ql==l&&r==qr) {
		if(seg[id].val.mn==1)
		return seg[id].val.cnt;
		else return 0;
	}
	down(id,l,r);
	int mid=l+r>>1;
	if(qr<=mid) return query(id<<1,l,mid,ql,qr);
	else if(ql>mid) return query(id<<1|1,mid+1,r,ql,qr);
	else return query(id<<1,l,mid,ql,mid)+query(id<<1|1,mid+1,r,mid+1,qr);
}
void solve() {
	int n;cin>>n;
	build(1,1,n);
	for(int i=1;i<=n;i++){
		adj[i].clear();
	}
	for(int i=1;i<n;i++){
		int u,v;cin>>u>>v;
		adj[u].push_back(v);
		adj[v].push_back(u);
	}
	int ans=0;
	for(int i=1;i<=n;i++){
		modify(1,1,n,1,i,1);
		for(auto &v:adj[i]){
			if(v<i){
				modify(1,1,n,1,v,-1);
			}
		}
		ans+=query(1,1,n,1,i);
	}
	cout<<ans<<"\n";
}
signed main() {
	close;
	int t;cin>>t;
	while(t--) 
	solve();
}

K. Happy Equation

打了个表发现a为奇数始终输出1

在x>p的时候我们可以发现因为a里面至少有一个2的因子，

2的x次方乘上一堆数必定能被2的p次方整除所以在x>p时我们求x的a次方%2的p次方为0的数就行了

在x<=p的时候枚举这个x 统计答案

在x>p的时候我们需要让x里面2的因数的数量cnt *a>=p 这样就能整除了

然后我们可以算出cnt最少需要多少个

假如说cnt至少需要2个 x是属于（p，2^p)的就相当于算 p-2^p之间有多少2的2次方(4)的倍数

除一下再减一下就好了

#include<bits/stdc++.h>
#define close std::ios::sync_with_stdio(false),cin.tie(0),cout.tie(0)
using namespace std;
typedef long long ll;
const ll MAXN =  3e5+7;
ll mod;
const ll inf = 0x3f3f3f3f;
#define int ll
ll _power(ll a,int b) {
	ll ans=1,res=a;
	while(b) {
		if(b&1)
			ans=ans*res%mod;
		res=res*res%mod;
		b>>=1;
	}
	return ans%mod;
}

void solve() {
	int a,p;
	cin>>a>>p;
	if(a%2==1) {
		cout<<1<<"\n";
		return;
	}
	mod=(1<<p);
	int ans=0;
	for(int x=2; x<=p; x+=2) { //统计x<=p的时候 合法的数量 
		if(_power(a,x)==_power(x,a)) ans++;
	}
	int k=p/a;
	if(k*a!=p) k++;
	ans+=pow(2,p-k);
	int l=p/pow(2,k);
	ans-=l;
	printf("%lld\n",ans);
	//cout<<ans-p/(pow(2,k))<<"\n";//此时必须满足x> p 
}
signed main() {
	int t;
	cin>>t;
	while(t--)
		solve();
}

2023暑期训练后半

1. 2023 Xian Jiaotong University Programming Contest

赛时过题ABCDEFHMN 补题GJO

2. 2019-2020 ACM-ICPC Brazil Subregional Programming Contest

3. 2019 ICPC Universidad Nacional de Colombia Programming

赛时过题BGIK 补CDJ

B. Boring Non-Palindrome

签到队友写的

void solve()
{
    //	freopen("data.in.txt","r",stdin);
    //	freopen("data.out.txt","w",stdout);
    cin >> s;
    int n = s.length();
    s = ' ' + s;
    for (int i = 1; i <= n; i++)
    {
        int l = i, r = n;
        while (l <= r)
        {
            if (s[l] == s[r])
            {
                l++, r--;
            }
            else
            {
                break;
            }
        }
        if (r < l)
        {
            cout << s;
            for (int j = i - 1; j > 0; j--)
                cout << s[j];
            return;
        }
    }
}

C. Common Subsequence

类似求最长匹配的dp 队友补了

D. Do Not Try This Problem

并查集+根号分治

间隔长的直接跳就行了短的用并查集暴力维护下一个跳的

int f[MAXN][320];//用并查集跳边 
int vis[MAXN];
struct node{
	int s,d,p;
	char ch;
}N[MAXN];
void clean(int n){
	for(int j=1;j<=n;j++)
	for(int i=1;i<320;i++){
		f[j][i]=j;
	}
}
int find(int x,int k){
	if(x!=f[x][k]) f[x][k]=find(f[x][k],k);
	return f[x][k];
}
void add(int x,int y,int k){
	int fx=find(x,k),fy=find(y,k);
	if(fx!=fy) f[fx][k]=fy;
}
void solve(){
	string str;cin>>str;
	int n=str.length();
	str=" "+str;
	int k=sqrt(n);
	clean(n);
	int q;cin>>q;
	for(int i=1;i<=q;i++){
		int s,d,p;cin>>N[i].s>>N[i].d>>N[i].p;
		char ch;cin>>N[i].ch;
	}
	for(int i=q;i>=1;i--){
		int s=N[i].s,d=N[i].d,p=N[i].p;
		char ch=N[i].ch;
		if(d>k){
			for(int j=s;j<=min(s+p*d,n);j+=d){
				if(!vis[j]){
					vis[j]=1;
					str[j]=ch;
				}
			}
		}
		else{
			int now=s;
			while(now<=min(s+p*d,n)){
				now=find(now,d);
				if(now>min(s+p*d,n)) break;
				if(!vis[now]){
					vis[now]=1;
					str[now]=ch;
					if(now+d<=n)
					add(now,now+d,d);
					now+=d;
				}
				else{
					if(now+d<=n)
					add(now,now+d,d);
					now+=d;
				}
			}
		}
	//	cout<<i<<str<<"\n";
	}
	for(int i=1;i<=n;i++) cout<<str[i];
}

E. Extreme Image

圆形的扫描线还没补

F. Fraction Formula

模拟

G. Graduation

struct E
{
    int to, w, next;
} edge[M];
int head[N], tot;
int sz[N];
int n, k;

void add(int u, int v, int w)
{
    tot++;
    edge[tot].to = v;
    edge[tot].w = w;
    edge[tot].next = head[u];
    head[u] = tot;
}

void dfs(int u, int f)
{
    sz[u] = 1;
    for (int i = head[u]; i; i = edge[i].next)
    {
        int v = edge[i].to, w = edge[i].w;
        dfs(v, u);
        sz[u] += sz[v];
    }
}

struct cmp
{
    bool operator()(int &a, int &b)
    {
        return sz[a] < sz[b];
    }
};

void solve()
{
    cin >> n >> k;
    vector<int> s;
    for (int i = 1, f; i <= n; i++)
    {
        cin >> f;
        if (f == 0)
            s.push_back(i);
        add(f, i, 0);
    }
    for (auto i : s)
        dfs(i, 0);
    priority_queue<int, vector<int>, cmp> q;
    for (auto i : s)
        q.push(i);
    int ans = 0;
    while (q.size())
    {
        vector<int> tp;
        for (int i = 1; q.size() && i <= k; i++)
        {
            int u = q.top();
            q.pop();
            for (int i = head[u]; i; i = edge[i].next)
            {
                tp.push_back(edge[i].to);
            }
        }
        for (auto i : tp)
            q.push(i);
        ans++;
    }
    cout << ans;
}

I. Integer Prefix

签到

void solve()
{
    string s;
    cin >> s;
    vector<char> ans;
    for (auto x : s)
    {
        if (x <= '9' && x >= '0')
            ans.push_back(x);
        else
            break;
    }
    if (ans.size() == 0)
        cout << -1 << '\n';
    else
        for (auto x : ans)
        {
            cout << x;
        }
}

J. Jail Destruction

砍墙势能线段树但是我势能用错了t了

要算非0的最小值这样最后是可以执行区间加的保持势能

#define int ll
int a[MAXN];
struct Info{
	int mn,sum,len;
};
Info operator +(const Info &a,const Info &b){
	Info c;
	c.len=a.len+b.len;
	c.mn=min(a.mn,b.mn);
	c.sum=a.sum+b.sum;
	return c;
}
struct node{
	int lazy;
	Info val;
}seg[MAXN<<2];
void settag(int id,int l,int r,int val){
	seg[id].val.mn+=val;
	seg[id].val.sum+=(seg[id].val.len)*val;
	seg[id].lazy+=val;
}
void up(int id){
	seg[id].val=seg[id<<1].val+seg[id<<1|1].val;
}
void down(int id,int l,int r){
	if(seg[id].lazy==0) return ;
	int mid=l+r>>1;
	settag(id<<1,l,mid,seg[id].lazy);
	settag(id<<1|1,mid+1,r,seg[id].lazy);
	seg[id].lazy=0;
}
void build(int id,int l,int r){
	if(l==r){
		seg[id].val.mn=a[l];
		seg[id].val.len=1;
		seg[id].val.sum=a[l];
		seg[id].lazy=0;
		return;
	}
	int mid=l+r>>1;
	build(id<<1,l,mid);
	build(id<<1|1,mid+1,r);
	up(id);
}
void modify(int id,int l,int r,int ql,int qr,int val){
	if(ql==l&&qr==r&&seg[id].val.mn+val>=0){
		settag(id,l,r,val);
		return;
	}
	if(seg[id].val.len==0) return;
	if(l==r){
		seg[id].val.sum=0;
		seg[id].val.mn=inf;
		seg[id].val.len=0;
		return;
	}
	down(id,l,r);
	int mid=l+r>>1;
	if(qr<=mid) modify(id<<1,l,mid,ql,qr,val);
	else if(ql>mid) modify(id<<1|1,mid+1,r,ql,qr,val);
	else{
		modify(id<<1,l,mid,ql,mid,val);
		modify(id<<1|1,mid+1,r,mid+1,qr,val);
	}
	up(id);
}
Info query(int id,int l,int r,int ql,int qr){
	if(ql==l&&r==qr){
		return seg[id].val;
	}
	down(id,l,r);
	int mid=l+r>>1;
	if(qr<=mid) return query(id<<1,l,mid,ql,qr);
	else if(ql>mid) return query(id<<1|1,mid+1,r,ql,qr);
	else{
		return query(id<<1,l,mid,ql,mid)+query(id<<1|1,mid+1,r,mid+1,qr);
	}
}
void solve()
{
	int n,m;cin>>n>>m;
	for(int i=1;i<=n;i++) cin>>a[i];
	build(1,1,n);
	for(int i=1;i<=m;i++){
		int op;cin>>op;
		if(op==1){
			int l,r;cin>>l>>r;
			if(r>=l) cout<<query(1,1,n,l,r).sum<<"\n";
			else{
				cout<<query(1,1,n,1,r).sum+query(1,1,n,l,n).sum<<"\n";
			}
		}
		else{
			int l,r,w;cin>>l>>r>>w;
			if(r>=l) modify(1,1,n,l,r,-w);
			else{
				modify(1,1,n,1,r,-w);
				modify(1,1,n,l,n,-w);
			}
		}
	}
}

K. Kernel Of Love

签到

void solve()
{
    int n;
    cin >> n;
    if (n == 1)
        cout << 0 << '\n';
    else if (n == 2)
        cout << 0 << '\n';
    else if (n == 3)
        cout << 2 << '\n';
    else if (n == 4)
        cout << 3 << '\n';
    else
        cout << (n - 4) / 3 * 2 + 3 + ((n - 4) % 3 >= 2) << '\n';
}

4. 2018-2019 9th BSUIR Open Programming Championship

赛时过题AEFH 补题IK

A. Four minutes until BSUIR Open

int n;
double a1, a2, x[N], v[N];

double count(double v0, double v1, double x)
{
    double a = (a1 * a1 + a1 * a2);
    double b = 2.0 * v0 * (a1 + a2);
    double c = v0 * v0 - v1 * v1 - 2.0 * a2 * x;
    // cout << a << ' ' << b << ' ' << c << endl;
    double t = (-b + sqrt(b * b - 4.0 * a * c)) / (2.0 * a);
    // cout << t + (v0 - v1 + a1 * t) / a2 << endl;
    return t + (v0 - v1 + a1 * t) / a2;
}

void solve()
{
    //	freopen("data.in.txt","r",stdin);
    //	freopen("data.out.txt","w",stdout);
    // cout << sqrt(6) << endl;
    cin >> n >> a1 >> a2;
    for (int i = 1; i <= n; i++)
        cin >> x[i] >> v[i];
    for (int i = n; i > 0; i--)
    {
        v[i - 1] = min(v[i - 1], sqrt(2.0 * a2 * (x[i] - x[i - 1]) + v[i] * v[i]));
    }
    for (int i = 0; i < n; i++)
    {
        v[i + 1] = min(v[i + 1], sqrt(2.0 * a1 * (x[i + 1] - x[i]) + v[i] * v[i]));
    }
    double ans = 0;
    for (int i = 1; i <= n; i++)
    {
        ans += count(v[i - 1], v[i], x[i] - x[i - 1]);
    }
    cout << fixed << setprecision(7) << ans << endl;
}

E. Small business

*Edgrass 搓了好久的模拟类目了居然给他过了

裱起来（

#include<bits/stdc++.h>
#define ll long long
#define PII pair<int,int>
//  #define int long long
using namespace std;
const ll mod=1e9+7; 
const ll inf=0x3f3f3f3f;  
const ll INF=0x3f3f3f3f3f3f3f3f;  
const double eps=1e-10;  
const int N=2e5+10;
// struct node{
//     friend bool operator<(const node&a,const node&b){
//         return ;
//     }
// }
inline int read(){int x=0,f=1;char ch=getchar();while(ch<'0'||ch>'9'){if(ch=='-')f=-1;ch=getchar();}while(ch>='0'&&ch<='9'){x=(x<<1)+(x<<3)+(ch^48);ch=getchar();}return x*f;}
inline void write(int x){char F[200];int tmp=x>0?x:-x;if(x<0) putchar('-');int cnt=0;while(tmp>0){F[cnt++]=tmp%10+'0';tmp/=10;}while(cnt>0)putchar(F[--cnt]);}
inline int combination(int n,int k){int sum=0;if (n==k||k==0){return 1;}else{return combination(n-1,k)+combination(n-1,k-1);}}

void solve(){
    string s;
    cin>>s;
    int n=s.size(),cnt0=0;
    vector<int>vec;
    for(auto x:s){
        if(x=='0')cnt0++;
        else vec.push_back(x-'0');
    }
    for(int i=1;i<=n;i++){
        if(i>n-i||(n-i)>19||i>19)continue;
        priority_queue<int,vector<int>,greater<int> >pq;
        for(auto x:vec){
            pq.push(x);
        }
        int cnt=cnt0,ok=1;
        vector<int>ans1,ans2;
        if(n-i==19){
            for(int i=1;i<=18;i++){
                cnt--;
            }
            if(pq.size()==0){
                    ok=0;
                    continue;
                }
            int t=pq.top();
            pq.pop();
            if(cnt<0||t!=1)continue;
            else {
                if(i==1){
                    if(cnt){
                        cnt--;
                        ans1.push_back(0);
                    }
                }
                if(ans1.size()!=i){
                    if(ok==0||pq.size()==0){
                        continue;
                    }
                    ans1.push_back(pq.top());
                    pq.pop();
                    for(int j=2;j<=i&&ok;j++){
                        if(cnt)ans1.push_back(0),cnt--;
                        else if(pq.size()==0){
                            ok=0;   
                            break;
                        }else ans1.push_back(pq.top()),pq.pop();
                    }
                }
                if((ans1[0]!=1||ans1[ans1.size()-1]!=0)&&ans1.size()==19)continue;
                for(auto x:ans1){
                    cout<<x;
                }
                cout<<" 1000000000000000000\n";
            }
            return ;
        }
        if(i==1){
            if(cnt){
                cnt--;
                ans1.push_back(0);
            }
        }
        if(n-i==1){
            if(cnt){
                cnt--;
                ans2.push_back(0);
            }
        }

        if(ans1.size()!=i){
            if(ok==0||pq.size()==0){
                continue;
            }
            ans1.push_back(pq.top());
            pq.pop();
            for(int j=2;j<=i&&ok;j++){
                if(cnt)ans1.push_back(0),cnt--;
                else if(pq.size()==0){
                    ok=0;   
                    break;
                }else ans1.push_back(pq.top()),pq.pop();
            }
        }

        if(ans2.size()!=n-i){
            if(ok==0||pq.size()==0){
                continue;
            }
            ans2.push_back(pq.top());
            pq.pop();
            for(int j=2;j<=(n-i);j++){
                if(cnt)ans2.push_back(0),cnt--;
                else if(pq.size()==0){
                    ok=0;   
                    break;
                }else ans2.push_back(pq.top()),pq.pop();
            }
            if(ok==0||pq.size()!=0||cnt!=0){
                continue;
            }
        }
        if((ans2[0]!=1||ans2[ans2.size()-1]!=0)&&ans2.size()==19)continue;
        if((ans1[0]!=1||ans1[ans1.size()-1]!=0)&&ans1.size()==19)continue;

        for(auto x:ans1){
            cout<<x;
        }
        cout<<" ";
        for(auto x:ans2){
            cout<<x;
        }
        cout<<'\n';
        return ;
    }
    cout<<-1<<" "<<-1<<'\n';
    //puts(ans>0?"YES":"NO");
    //puts(ans>0?"Yes":"No");
}

F. Prime or number

void solve()
{
    int n;
    cin >> n;
    if (n == 1)
    {
        cout << "No\n";
        return;
    }
    int x = n, cnt = 0;
    while (x)
    {
        if (x % 2)
            cnt++;
        x /= 2;
    }
    if (cnt == 1)
        cout << "Yes\n";
    else if (cnt == 2 && n % 2)
        cout << "Yes\n";
    else
        cout << "No\n";
    //	freopen("data.in.txt","r",stdin);
    //	freopen("data.out.txt","w",stdout);
}

H. Nonfibonacci numbers

有几个数字固定不会出现数位dp

int f[N][N][2][2], num[N], now;
int dfs(int pos, int sum, bool lead, bool limit)
{
    int ans = 0;
    if (pos == 0)
        return 1;
    if (f[pos][sum][limit][lead] != -1)
        return f[pos][sum][limit][lead];
    int up = (limit ? num[pos] : 9);
    // cout<<"pos:"<<pos<<" "<<up<<'\n';
    for (int i = 0; i <= up; i++)
    {
        if (i == 1 || i == 2 || i == 3 || i == 5 || i == 8)
            continue;
        if (i == 0 && lead)
            ans += dfs(pos - 1, sum, true, limit &&i == up);
        else
            ans += dfs(pos - 1, sum + 1, false, limit &&i == up);
    }
    f[pos][sum][limit][lead] = ans;
    return ans;
}
void solve()
{
    int n=read();
    int x = n, len = 0;
    while (x)
    {
        num[++len] = x % 10;
        x /= 10;
    }
    memset(f, -1, sizeof f);
    cout << dfs(len, 0, true, true) << '\n';
    //    freopen("data.in.txt","r",stdin);
    //    freopen("data.out.txt","w",stdout);
}

I. Equal Mod Segments

求连续mod 从左到右和从右到左一样的区间个数

用二分求对每个数来说向下取的区间

线段树补充 - xishuiw - 博客园 (cnblogs.com)

丢这了扫描线

K. Innovations

垃圾做法树剖但是赛场上线段树板子抄错了

ll a[MAXN];
vector<pair<int,ll> > adj[MAXN];
int deep[MAXN],siz[MAXN],hson[MAXN],rak[MAXN],dfn[MAXN],f[MAXN],cnt=0,top[MAXN];
ll ans=0,sum;
void tree_build(int u,int fa,int dep) {
	deep[u]=dep;
	siz[u]=1;
	f[u]=fa;
	for(auto &it:adj[u]) {
		int v=it.first;
		if(v==fa) continue;
		ll w=it.second;
		a[v]=w;
		tree_build(v,u,dep+1);
		siz[u]+=siz[v];
		if(hson[u]==0||siz[v]>siz[hson[u]]) hson[u]=v;
	}
}
void tree_decom(int u,int t) {
	top[u]=t;
	cnt++;
	dfn[u]=cnt;
	rak[cnt]=u;
	if(hson[u]!=0) {
		tree_decom(hson[u],t);
		for(auto &it:adj[u]) {
			int v=it.first;
			if(hson[u]!=v&&v!=f[u]) tree_decom(v,v);
		}
	}
}
struct Info {
	ll sum;
	bool flag;
};
Info operator +(const Info &a,const Info &b) {
	Info c;
	c.flag=(a.flag&b.flag);
	return c;
}
struct node {
	Info val;
} seg[MAXN<<2];
void up(int id) {
	seg[id].val=seg[id<<1].val+seg[id<<1|1].val;
}
void build(int id,int l,int r) {
	if(l==r) {
		seg[id].val.sum=a[rak[l]];
		if(a[rak[l]]==1) seg[id].val.flag=1;
		return;
	}
	int mid=l+r>>1;
	build(id<<1,l,mid);
	build(id<<1|1,mid+1,r);
	up(id);
}
void modify(int id,int l,int r,int ql,int qr,int val) {
	if(seg[id].val.flag==1) return;
	if(ql==l&&r==qr) {
		if(seg[id].val.flag==1) {
			return;
		}
	}
	if(l==r) {
		ll tmp=sqrt(seg[id].val.sum);
		ll temp=seg[id].val.sum-tmp;
		int sizz=siz[rak[l]];
		ans-=(temp)*(sizz)*(sum-sizz);
		ans%=mod;
		if(ans<0) ans+=mod;
		seg[id].val.sum=tmp;
		if(tmp==1) seg[id].val.flag=1;
		return;
	}
	int mid=(l+r)>>1;
	if(qr<=mid) modify(id<<1,l,mid,ql,qr,val);
	else if(ql>mid) modify(id<<1|1,mid+1,r,ql,qr,val);
	else {
		if(seg[id<<1].val.flag!=1)
		modify(id<<1,l,mid,ql,mid,val);
		if(seg[id<<1|1].val.flag!=1)
		modify(id<<1|1,mid+1,r,mid+1,qr,val);
	}
	up(id);
}
void dfs(int u,int fa) {
	for(auto &it:adj[u]) {
		int v=it.first;
		if(v==fa) continue;
		ans+=a[v]*(siz[v])*(sum-siz[v]);
		if(ans>mod)
		ans%=mod;
		dfs(v,u);
	}
}
void solve() {
	int n,q;
	scanf("%d %d",&n,&q);
	sum=n;
	for(int i=1; i<n; i++) {
		int u,v;
		ll w;
		scanf("%d %d %lld",&u,&v,&w);
		adj[u].push_back({v,w});
		adj[v].push_back({u,w});
	}
	tree_build(1,0,0);
	tree_decom(1,1);
	dfs(1,0);
	build(1,1,n);
	printf("%lld\n",ans);
	for(int i=1; i<=q; i++) {
		int u,v;
		scanf("%d %d",&u,&v);
		if(u==v||seg[1].val.flag==1) {
			printf("%lld\n",ans);
			continue;
		}
		while(top[u]!=top[v]) {
			if(deep[top[u]]>deep[top[v]]) {
				modify(1,1,n,dfn[top[u]],dfn[u],1);
				u=f[top[u]];
			} else {
				modify(1,1,n,dfn[top[v]],dfn[v],1);
				v=f[top[v]];
			}
		}
		if(dfn[v]!=dfn[u]) {
			if(dfn[v]<dfn[u]) swap(u,v);
			modify(1,1,n,dfn[u]+1,dfn[v],1);
		}

		printf("%lld\n",ans);
	}
}

5. 2019-2020 Winter Petrozavodsk Camp, Day 8: Almost Algorithmic Contest

只写了FG的超难场

6. 2020-2021 ACM-ICPC Brazil Subregional Programming Contest

赛时过题：ABFGL 补题：EHN（I）

A. Sticker Album

概率dp 一直想不明白怎么搞的样例然后发现从0开始推

用前缀和（滑动窗口）优化速度

对于a==0的样例xlen/（len-）就行了

半蒙的

void solve() {
	int n,a,b;
	cin>>n>>a>>b;
	
	for(int i=1; i<=n; i++) {
		int l=i-b;
		int r=i-a;
		if(i<=a) dp[i]=1;
		else {
			if(l>0){
				if(a!=0)
				dp[i]=(pre[r]-pre[l-1]+r-l+1)/(b-a+1);
				else{
					dp[i]=(pre[r-1]-pre[l-1]+(r-l+1))/(b-a+1);
					dp[i]*=(b-a+1)*1.0/(b-a);
				}
			}
			else{
				if(a!=0)
				dp[i]=(pre[r]+r-l+1)/(b-a+1);
				else{
					dp[i]=(pre[r-1]+r-l+1)/(b-a+1);
					dp[i]*=(b-a+1)*1.0/(b-a);
				}
			}
		}

		pre[i]=pre[i-1]+dp[i];
	//	cout<<"i "<<i<<" :"<<dp[i]<<"\n";
	}
	printf("%.8f",dp[n]);
}

B. Battleship

队友签到的

int mp[20][20], n;
 
void solve()
{
    //	freopen("data.in.txt","r",stdin);
    //	freopen("data.out.txt","w",stdout);
    cin >> n;
    for (int i = 1; i <= n; i++)
    {
        int d, l, r, c;
        cin >> d >> l >> r >> c;
        for (int i = 0; i < l; i++)
        {
            int x = r, y = c;
            if (d == 0)
                y += i;
            else
                x += i;
            if (x > 10 || y > 10 || mp[x][y] == 1)
            {
                cout << "N";
                return;
            }
            mp[x][y] = 1;
        }
    }
    cout << "Y";
}

E. Party Company

扫描线赛场没写出来

树状数组维护符合的下界数量

int t[MAXN];
int lowbit(int x) {
	return x&-x;
}
ll getsum(int x) {
	ll sum=0;
	while(x) {
 
		sum+=t[x];
		x-=lowbit(x);
	}
	return sum;
}
void addv(int x,int val) {
	while(x<=MAXN) {
		t[x]+=val;
		x+=lowbit(x);
	}
}
struct E {
	int to,w,next;
} edge[M];
int head[N],tot;
 
void add(int u,int v,int w) {
	tot++;
	edge[tot].to=v;
	edge[tot].w=w;
	edge[tot].next=head[u];
	head[u]=tot;
}
int fa[N][20];
int w[MAXN];
vector<int> node[MAXN];
int ans[MAXN],far[MAXN];
void dfs(int u,int f) {
	fa[u][0]=f;
	for(int i=1; i<20; i++)fa[u][i]=fa[fa[u][i-1]][i-1];
 
	for(int i=head[u]; i; i=edge[i].next) {
		int v=edge[i].to,w=edge[i].w;
		if(v==f)continue;
		dfs(v,u);
	}
}
int gettop(int u,int mx) {
	for(int i=19; i>=0; i--) {
		if(w[fa[u][i]]<=mx)u=fa[u][i];
	}
	return u;
}
 
void calc(int u,int fa) {
	for(auto &it:node[u]) {
		addv(it,1);
	}
	ans[u]=getsum(w[u]);
	for(int i=head[u]; i; i=edge[i].next) {
		int v=edge[i].to;
		if(v==fa) continue;
		calc(v,u);
	}
	for(auto &it:node[u]) {
		addv(it,-1);
	}
}
void solve() {
	int n,m;
	cin>>n>>m;
	for(int i=1; i<=n; i++) {
		int u;
		cin>>w[i]>>u;
		if(i!=1)
			add(i,u,1);
		add(u,i,1);
		far[i]=u;
	}
	far[1]=0;
	dfs(1,1);
	for(int i=1; i<=m; i++) {
		int u,l,r;
		cin>>u>>l>>r;
		int top=gettop(u,r);
		node[top].push_back(l);
	}
	calc(1,1);
	for(int i=1; i<=n; i++) cout<<ans[i]<<" ";
}

F. Fastminton

模拟

string s;
 
void solve()
{
    //	freopen("data.in.txt","r",stdin);
    //	freopen("data.out.txt","w",stdout);
    int ls = 0, rs = 0, lcnt = 0, rcnt = 0, tag = 0;
 
    cin >> s;
    // bool flag=true;
    for (int i = 0; i < s.length(); i++)
    {
 
        if (s[i] == 'Q')
        {
            if (ls == 2)
                printf("2 (winner) - %d\n", rs);
            else if (rs == 2)
                printf("%d - 2 (winner)\n", ls);
            else if (tag == 0)
                printf("%d (%d*) - %d (%d)\n", ls, lcnt, rs, rcnt);
            else
                printf("%d (%d) - %d (%d*)\n", ls, lcnt, rs, rcnt);
        }
        if (ls == 2 || rs == 2)
            continue;
 
        if (s[i] == 'S')
        {
            if (tag == 0)
                lcnt++;
            else
                rcnt++;
        }
        else if (s[i] == 'R')
        {
            tag = 1 - tag;
            if (tag == 0)
                lcnt++;
            else
                rcnt++;
        }
        if (lcnt == 10 || lcnt >= 5 && lcnt >= rcnt + 2)
        {
            ls++;
            lcnt = rcnt = 0;
        }
        else if (rcnt == 10 || rcnt >= 5 && rcnt >= lcnt + 2)
        {
            rs++;
            lcnt = rcnt = 0;
        }
    }
}

G. Game Show

最大前缀和签到

H. SBC's Hangar

每个数倍数差都在2以上

意思就是从小到大排序后大的那个的数量小的所有加起来都取不到

用二进制表示是否取这个

因此可以输出最大的能表示的范围内的二进制情况

然后也能得到最小的满足范围下界的二进制

对其中的式子进行数位dp

#define int ll
int a[MAXN];
int upper[MAXN];int n,k;
int dp[60][2][60];
ll dfs(int pos,int pre_num,int c,int flag){
	int max_number;
	if(pos<=0) return c==k;
	if(!flag&&dp[pos][pre_num][c]!=-1) return dp[pos][pre_num][c];
	if(flag) max_number=upper[pos];
	else max_number=1;
	ll ret=0;
	for(int i=0;i<=max_number;i++){

		ret+=dfs(pos-1,i,c+i,flag&&(i==max_number));
		
	}
	if(!flag) dp[pos][pre_num][c]=ret;
	return ret;
}
void solve(){
	cin>>n>>k;
	for(int i=1;i<=n;i++) cin>>a[i];
	sort(a+1,a+1+n);
	int l,r;cin>>l>>r;
	int tmp=r;
	for(int i=n;i>=1;i--){
		if(a[i]<=tmp){
			upper[i]=1;
			tmp-=a[i];
		}
		else upper[i]=0;
	}
	for(int i=0;i<=n;i++){
		for(int j=0;j<=1;j++){
			for(int x=0;x<=n;x++) dp[i][j][x]=-1;
		}
	}
	int ans=dfs(n,0,0,1);
	tmp=0;
	int cnt=0;
	for(int i=n;i>=1;i--){
		int temp=0;
			for(int j=1;j<i;j++){
				temp+=a[j];
			}
		if(temp+tmp<l){
			tmp+=a[i];
			upper[i]=1;
			cnt++;
		}
		else upper[i]=0;
	}
	for(int i=0;i<=n;i++){
		for(int j=0;j<=1;j++){
			for(int x=0;x<=n;x++) dp[i][j][x]=-1;
		}
	}
	if(cnt==k&&tmp>=l) ans++;
	ans-=dfs(n,0,0,1);
	cout<<ans;
}

I. Interactivity

树dp 队友式子推对了但是用逆元的组合数就错了

N. Number Multiplication

套了个n1/4的分解质因数的板子就过了

就板子不贴了毛的

7. 2018-2019 ACM-ICPC Brazil Subregional Programming Contest

赛时CDEFIL 补题BG

B. Marbles

博弈

C. Pizza Cutter

扫描线就逆序对那个

如果加了一条线多了a个交点那么方块的个数多a+1

扫两遍就行

因为只有Li>Lj&&Ri<Rj的才有交点很好统计

#define int ll
int t[MAXN];
int lowbit(int x) {
	return x&-x;
}
ll getsum(int x) {
	ll sum=0;
	while(x) {
		sum+=t[x];
		x-=lowbit(x);
	}
	return sum;
}
void addv(int x,int val) {
	while(x<MAXN) {
		t[x]+=val;
		x+=lowbit(x);
	}
}
vector<int> X;
int getX(int x) {
	int k=lower_bound(X.begin(),X.end(),x)-X.begin();
	return k;
}
struct node{
	int x,y;
}N[MAXN];
bool bj(node a,node b){
	return a.x>b.x;
}
void solve() {
	int x,y;
	cin>>x>>y;
	int n,m;cin>>n>>m;
	for(int i=1;i<=n;i++){
		cin>>N[i].x>>N[i].y;
		X.push_back(N[i].x);
		X.push_back(N[i].y);
	}
	X.push_back(inf*-1);
	sort(X.begin(),X.end());
	X.erase(unique(X.begin(),X.end()),X.end());
	int ans=1;
	sort(N+1,N+1+n,bj);
	for(int i=1;i<=n;i++){
		int a=N[i].x;
		int b=N[i].y;
		b=getX(b);
		ans+=1+getsum(b);
		addv(b,1);
	}
	memset(t,0,sizeof(t));
	X.clear();
	for(int i=1;i<=m;i++){
		cin>>N[i].x>>N[i].y;
		X.push_back(N[i].x);
		X.push_back(N[i].y);
	}
	X.push_back(inf*-1);
	sort(X.begin(),X.end());
	X.erase(unique(X.begin(),X.end()),X.end());
	sort(N+1,N+1+m,bj);
	for(int i=1;i<=m;i++){
		int a=N[i].x;
		int b=N[i].y;
		b=getX(b);
		ans+=1+getsum(b);
		addv(b,1);
	}
	ans+=n*m;
	cout<<ans;
}

D. Unraveling Monty Hall

队友签到的


void solve()
{
    //	freopen("data.in.txt","r",stdin);
    //	freopen("data.out.txt","w",stdout);
    int n, a, ans = 0;
    cin >> n;
    while (n--)
    {
        cin >> a;
        if (a > 1)
            ans++;
    }
    cout << ans;
}

E. Enigma

对每个字母扫它的贡献值算在字符串的头上

计算有几个字符串位置的头贡献是0就行了

int len[MAXN];
void solve()
{
   string a,b;cin>>a>>b;
   int na=a.length(),nb=b.length();
   b=" "+b;
   a=" "+a;
   int st=na-nb;
   for(int i=1;i<=nb;i++){
   		for(int j=i;j<=st+i;j++){
   			if(a[j]==b[i]){
   				len[j-i+1]++;
   		//		cout<<b[i]<<" "<<j-i+1<<"\n";
			   }
		   }
   }
   int ans=0;
   for(int i=1;i<=st+1;i++){
   	if(len[i]==0) ans++;
   }
   cout<<ans;
}

F. Music Festival

每个舞会至少选一个人表演时间不能重合求最大贡献

状压dp

vector<array<int,3> >g[86405];
int f[86405][1100];
void solve(){
    int n=read();
    for(int i=1;i<=n;i++){
        int k=read();
        for(int j=1;j<=k;j++){
            int x=read(),y=read(),w=read();
            g[y].push_back((array<int,3>){x,w,i});
        }
    }
    for(int i=1;i<=86400;i++){
        for(int j=0;j<(1<<(n));j++){
            f[i][j]=f[i-1][j];
            for(auto it:g[i]){
                int l=it[0],w=it[1],k=it[2]-1;
                if(((j>>k)&1)&&(f[l][j-(1<<k)]>0))f[i][j]=max(f[i][j],f[l][j-(1<<k)]+w);
                if((((j&(((1<<n)-1)-(1<<k)))==0))||(f[l][(j&(((1<<n)-1)-(1<<k)))]>0))f[i][j]=max(f[i][j],f[l][(j&(((1<<n)-1)-(1<<k)))]+w);
            }
        }
    }
    // cerr<<1<<'\n';
    int ans=0;
    for(int i=1;i<=86400;i++){
        ans=max(ans,f[i][(1<<(n))-1]);
    }
    if(ans==0)cout<<-1<<'\n';
    else cout<<ans<<'\n';
    //puts(ans>0?"YES":"NO");
    //puts(ans>0?"Yes":"No");
}

G. Gasoline

二分+网络流

不会写队友补了

I. Switches

最多做两边就知道结果了

模拟一下

int aim[N], n, m, p[N];
vector<int> a[N];

bool check()
{
    for (int i = 1; i <= m; i++)
    {
        if (aim[i] != p[i])
            return false;
    }
    return true;
}

void solve()
{
    //	freopen("data.in.txt","r",stdin);
    //	freopen("data.out.txt","w",stdout);
    cin >> n >> m;
    int l;
    cin >> l;
    for (int i = 1; i <= l; i++)
    {
        int tp;
        cin >> tp;
        aim[tp] = 1;
    }
    for (int i = 1; i <= n; i++)
    {
        int num, tp;
        cin >> num;
        for (int j = 1; j <= num; j++)
        {
            cin >> tp;
            a[i].push_back(tp);
        }
    }
    int ans = 0;
    for (int i = 1; i <= n * 2; i++)
    {
        int id = (i - 1) % n + 1;
        for (auto pos : a[id])
            p[pos] ^= 1;
        ans++;
        if (check())
        {
            cout << ans << endl;
            return;
        }
    }
    // for (int i = 1; i <= m; i++)
    //     cout << p[i] << ' ';
    // cout << endl;
    cout << -1 << endl;
}

J. Joining Capitals

听说是最小斯坦纳树先学学

L. Subway Lines

树剖区间0/1翻转统计全数1的个数就可以算出变成0的个数

听讲题发现做麻烦了

using namespace std;
#define int ll
int deep[MAXN],siz[MAXN],f[MAXN],hson[MAXN],dfn[MAXN],top[MAXN],cnt=0,rdfn[MAXN],rak[MAXN];
vector<int> adj[MAXN];
struct Info{
	int sum;
};
Info operator +(const Info &a,const Info &b){
	Info c;
	c.sum=a.sum+b.sum;
	return c;
};
struct node{
	int lazy;
	Info val;
}seg[MAXN<<2];
void up(int id){
	seg[id].val=seg[id<<1].val+seg[id<<1|1].val;
}
void settag(int id,int l,int r,int tag){
	seg[id].val.sum=r-l+1-seg[id].val.sum;
	seg[id].lazy^=1;
}
void down(int id,int l,int r){
	if(seg[id].lazy==0) return;
	int mid=l+r>>1;
	settag(id<<1,l,mid,seg[id].lazy);
	settag(id<<1|1,mid+1,r,seg[id].lazy);
	seg[id].lazy=0;
}
void build(int id,int l,int r){
	if(l==r){
		seg[id].val.sum=0;
		seg[id].lazy=0;
		return;
	}
	int mid=l+r>>1;
	build(id<<1,l,mid);
	build(id<<1|1,mid+1,r);
	up(id);
}
void modify(int id,int l,int r,int ql,int qr,int val){
	if(ql==l&&r==qr){
		settag(id,l,r,val);
		return;
	}
	down(id,l,r);
	int mid=l+r>>1;
	if(qr<=mid) modify(id<<1,l,mid,ql,qr,val);
	else if(ql>mid) modify(id<<1|1,mid+1,r,ql,qr,val);
	else{
		modify(id<<1,l,mid,ql,mid,val);
		modify(id<<1|1,mid+1,r,mid+1,qr,val);
	}
	up(id);
}
Info query(int id,int l,int r,int ql,int qr){
	if(ql==l&&r==qr) return seg[id].val;
	down(id,l,r);
	int mid=l+r>>1;
	if(qr<=mid) return query(id<<1,l,mid,ql,qr);
	else if(ql>mid) return query(id<<1|1,mid+1,r,ql,qr);
	else{
		return query(id<<1,l,mid,ql,mid)+query(id<<1|1,mid+1,r,mid+1,qr);
	}
}
void tree_build(int u,int fa,int dep){
	deep[u]=dep;
	siz[u]=1;
	f[u]=fa;
	hson[u]=0;
	for(auto &v:adj[u]){
		if(v==fa) continue;
		tree_build(v,u,dep+1);
		siz[u]+=siz[v];
		if(hson[u]==0||siz[v]>siz[hson[u]]) hson[u]=v;
	}
}
void tree_decom(int u,int t){
	top[u]=t;
	cnt++;
	dfn[u]=cnt;
	rak[cnt]=u;
	if(hson[u]!=0){
		tree_decom(hson[u],t);
		for(auto &v:adj[u]){
			if(hson[u]!=v&&v!=f[u]) tree_decom(v,v);
		}
	}
	rdfn[u]=cnt;
}
int getlca(int u,int v){
	while(top[u]!=top[v]){
		if(deep[top[u]]>deep[top[v]]) u=f[top[u]];
		else v=f[top[v]];
	}
	return deep[u]>deep[v]?v:u;
}int n,m;
void calc(int u,int v){
		while(top[u]!=top[v]) {
			
			if(deep[top[u]]>deep[top[v]]) {
				modify(1,1,n,dfn[top[u]],dfn[u],1);
				u=f[top[u]];
			} else {
				modify(1,1,n,dfn[top[v]],dfn[v],1);
				v=f[top[v]];
			}
		}
			if(dfn[v]<dfn[u]) swap(u,v);		
			modify(1,1,n,dfn[u],dfn[v],1);
//		cout<<"* ";
//		for(int i=1;i<=n;i++){
//			cout<<query(1,1,n,i,i).sum<<" ";
//		}
//		cout<<"\n";
}
void solve()
{
	cin>>n>>m;
	for(int i=1;i<n;i++){
		int u,v;cin>>u>>v;
		adj[u].push_back(v);
		adj[v].push_back(u);
	}
	tree_build(1,0,0);
	tree_decom(1,1);
//	for(int i=1;i<=n;i++) cout<<dfn[i]<<" ";
//		for(int i=1;i<=n;i++) cout<<top[i]<<" ";
//	cout<<"\n";
	build(1,1,n);
	while(m--){
		int u,v,p,q;
		cin>>u>>v;
		calc(u,v);
		int len1=deep[u]+deep[v]-deep[getlca(u,v)]*2+1;
		cin>>p>>q;
		calc(p,q);
		int len2=deep[p]+deep[q]-deep[getlca(p,q)]*2+1;
		int ans=len1+len2-query(1,1,n,1,n).sum;
		ans/=2;
		cout<<ans<<'\n';
		calc(u,v);
		calc(p,q);
	}
   
}

M. Modifying SAT

赛时写的2-sat 好像是高斯消元晚点补

8. 2020-2021 Winter Petrozavodsk Camp, Day 5: Almost Retired Dandelion Contest (XXI Open Cup, Grand Prix of Nizhny Novgorod)

赛时过题:GJM 补题A

A. Assignment Problem

回溯法暴力搜索

赛场上没算明白不敢写

#include<bits/stdc++.h>
#define close std::ios::sync_with_stdio(false),cin.tie(0),cout.tie(0)
using namespace std;
typedef long long ll;
const ll MAXN =  3e5+7;
const ll mod = 1e9+7;
const ll inf = 0x3f3f3f3f;
int a[20][1050];
int f[MAXN];
int ans[MAXN];
int vis[MAXN];
int n,m;
void dfs(int p){
	int q=0;
	for(int i=1;i<=m;i++){
		if(f[a[p][i]]==0){
			q=a[p][i];
			break;
		}
	}
	f[q]=1;
	ans[q]=1;
	for(int i=1;i<=m;i++){
		if(vis[i]) continue;
		vis[i]=1;
		dfs(i);
		vis[i]=0;
	}
	f[q]=0;
	return;
}
void solve(){
	cin>>n>>m;
	for(int i=1;i<=m;i++){
		for(int j=1;j<=n;j++) cin>>a[i][j];
	}
	for(int i=1;i<=m;i++){
		vis[i]=1;
		dfs(i);
		vis[i]=0;
	}
	int cnt=0;
	for(int i=1;i<=n;i++){
		if(ans[i]) cnt++;
	}
	cout<<cnt<<"\n";
	for(int i=1;i<=n;i++){
		if(ans[i])
		cout<<i<<" ";
	}
}

G. Remove the Prime

用n1/4分解质因数然后nim博弈代码太长了不丢了

J. Increasing or Decreasing

先给它整体排序一遍然后再从一遍开始一个一个对其刚好n次


int n;
int a[N], p[N];
int pos[N];

struct ans
{
    int l, r;
    char ch;
} as[N];
int cnt = 0;

int cmp1(int &a, int &b)
{
    return a > b;
}

void solve()
{
    //	freopen("data.in.txt","r",stdin);
    //	freopen("data.out.txt","w",stdout);
    cin >> n;
    for (int i = 1; i <= n; i++)
        cin >> a[i];
    as[++cnt] = {1, n, 'I'};
    sort(a + 1, a + 1 + n);
    for (int i = 1; i <= n; i++)
        pos[a[i]] = i;
    for (int i = 1; i <= n; i++)
        cin >> p[i];

    for (int i = n; i > 0; i--)
    {
        if (a[i] == p[i])
            continue;
        cnt++;
        int l = pos[p[i]], r = i;
        // cerr << l << ' ' << r << endl;
        sort(a + l, a + 1 + r);

        as[cnt].l = l;
        as[cnt].r = r;

        if (a[r] == p[i])
            as[cnt].ch = 'I';
        else
        {
            sort(a + l, a + 1 + r, cmp1);
            as[cnt].ch = 'D';
        }
        for (int j = l; j <= r; j++)
            pos[a[j]] = j;
        // for (int i = 1; i <= n; i++)
        //     cout << a[i] << ' ';
        // cout << endl;
    }
    cout << cnt << endl;
    for (int i = 1; i <= cnt; i++)
        cout << as[i].l << ' ' << as[i].r << ' ' << as[i].ch << endl;
}

M. Discrete Logarithm is a Joke

从最后一个数往前面跑 mod别写1e18+31 好像会爆ll

#include<bits/stdc++.h>
using namespace std;
#define int __int128
typedef long long ll;
const long long mod = 1000000000000000031;
int _power(int a,int b) {
	int ans=1,res=a;
	while(b) {
		if(b&1)
			ans=ans*res%mod;
			res=res*res%mod;
			b>>=1;
	}
	return ans%mod;
}
void cal() {
	long long n;
	cin>>n;
	long long st=300;
	for(int i=1000000-1; i>=n; i--) {
		st=_power(42,st);
	}
    cout<<st<<'\n';
}

posted @ 2023-08-17 02:08 xishuiw 阅读(86) 评论(1) 编辑收藏举报

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