Codeforces Round #447 (Div. 2)

A.记录每个A前面有多少个Q 后面有多少个Q

#include <bits/stdc++.h>
#define PI acos(-1.0)
#define mem(a,b) memset((a),b,sizeof(a))
#define TS printf("!!!\n")
#define pb push_back
#define inf 1e9
//std::ios::sync_with_stdio(false);
using namespace std;
//priority_queue<int,vector<int>,greater<int>> que;
const double EPS = 1.0e-8;
const double eps = 1.0e-8;
typedef pair<int, int> pairint;
typedef long long ll;
typedef unsigned long long ull;
const int maxn = 1e6 + 10;
const int  maxm = 300;
const int turn[4][2] = {{1, 0}, { -1, 0}, {0, 1}, {0, -1}};
//next_permutation
int a[maxn];
int b[maxn];
int n, now, cur, pop;
int main()
{
        //freopen("in.txt", "r", stdin);
        int left, right;
        string a1;
        cin >> a1;
        int len = a1.size();
        for (int i = len - 1; i >= 0; i--)
        {
                if (a1[i] == 'Q')
                {
                        a[i] = a[i + 1] + 1;
                }
                else
                {
                        a[i] = a[i + 1];
                }
        }
        for (int i = 0; i < len; i++)
        {
                if (a1[i] == 'Q')
                {
                        if (i == 0)
                        {
                                b[i] = 1;
                        }
                        else
                        {
                                b[i] = b[i - 1] + 1;
                        }
                }
                else
                {
                        if (i == 0)
                        {
                                b[i] = 0;
                        }
                        else
                        {
                                b[i] = b[i - 1];
                        }
                }
        }
        int anser = 0;
        for (int i = 0; i < len; i++)
        {
                if (a1[i] == 'A')
                {
                        anser += a[i] * b[i];
                }
        }
        cout << anser << endl;
}

B.直接快速幂(n-1)*(m-1)会炸 快速幂两次/欧拉函数降幂

快速幂两次:

#include <bits/stdc++.h>
#define PI acos(-1.0)
#define mem(a,b) memset((a),b,sizeof(a))
#define TS printf("!!!\n")
#define pb push_back
#define inf 1e9
//std::ios::sync_with_stdio(false);
using namespace std;
//priority_queue<int,vector<int>,greater<int>> que;
const double EPS = 1.0e-8;
const double eps = 1.0e-8;
typedef pair<int, int> pairint;
typedef long long ll;
typedef unsigned long long ull;
const int maxn = 1e6 + 10;
const int  maxm = 300;
const int turn[4][2] = {{1, 0}, { -1, 0}, {0, 1}, {0, -1}};
//next_permutation
ll mod = 1e9 + 7;
ll mod2 = mod - 1;
ll quick(ll a, ll b, ll mod)
{
        ll now = 1;
        while (b)
        {
                if (b % 2)
                {
                        now = (now * a) % mod;
                }
                a = (a * a) % mod;
                b >>= 1;
        }
        return now;
}
int main()
{
        ll n, m, k;
        cin >> n >> m >> k;
        if ((n == 1) || (m == 1))
        {
                if (k == 1)
                {
                        cout << 1 << endl;
                        return 0;
                }
                else
                {
                        if ((n + m) % 2)
                        {
                                cout << 0 << endl;
                                return 0;
                        }
                        else
                        {
                                cout << 1 << endl;
                                return 0;
                        }
                }
        }
        ll ans = quick(2, n - 1, mod);
        ans = quick(ans, m - 1, mod);
        if (k == 1)
        {
                cout << ans << endl;
        }
        else
        {
                if ((n + m) % 2)
                {
                        cout << 0 << endl;
                }
                else
                {
                        cout << ans << endl;
                }
        }
        return 0;
}

欧拉函数降幂:

#include <bits/stdc++.h>
#define PI acos(-1.0)
#define mem(a,b) memset((a),b,sizeof(a))
#define TS printf("!!!\n")
#define pb push_back
#define inf 1e9
//std::ios::sync_with_stdio(false);
using namespace std;
//priority_queue<int,vector<int>,greater<int>> que;
const double EPS = 1.0e-8;
const double eps = 1.0e-8;
typedef pair<int, int> pairint;
typedef long long ll;
typedef unsigned long long ull;
const int maxn = 1e6 + 10;
const int  maxm = 300;
const int turn[4][2] = {{1, 0}, { -1, 0}, {0, 1}, {0, -1}};
//next_permutation
ll mod = 1e9 + 7;
ll mod2 = mod - 1;
ll quick(ll a, ll b, ll mod)
{
        ll now = 1;
        while (b)
        {
                if (b % 2)
                {
                        now = (now * a) % mod;
                }
                a = (a * a) % mod;
                b >>= 1;
        }
        return now;
}
int main()
{
        ll n, m, k;
        cin >> n >> m >> k;
        if ((n == 1) || (m == 1))
        {
                if (k == 1)
                {
                        cout << 1 << endl;
                        return 0;
                }
                else
                {
                        if ((n + m) % 2)
                        {
                                cout << 0 << endl;
                                return 0;
                        }
                        else
                        {
                                cout << 1 << endl;
                                return 0;
                        }
                }
        }
        ll ans = quick(2, ((n - 1) % mod2) * ((m - 1) % mod2), mod);
        if (k == 1)
        {
                cout << ans % mod << endl;
        }
        else
        {
                if ((n + m) % 2)
                {
                        cout << 0 << endl;
                }
                else
                {
                        cout << ans % mod << endl;
                }
        }
        return 0;
}

C:毒瘤..... 一开始的做法错误 并不是N^2检验是否全部都存在然后输出原序列就行

因为他给你的数是按得到的数的大小递增排序给出 N^2检验 两个不一定是相连的

其实只要所有数的gcd等于最小的就可以 然后再每两个数之间插入最小的数 就构造成功了

#include <bits/stdc++.h>
#define PI acos(-1.0)
#define mem(a,b) memset((a),b,sizeof(a))
#define TS printf("!!!\n")
#define pb push_back
#define inf 1e9
//std::ios::sync_with_stdio(false);
using namespace std;
//priority_queue<int,vector<int>,greater<int>> que;
const double EPS = 1.0e-8;
const double eps = 1.0e-8;
typedef pair<int, int> pairint;
typedef long long ll;
typedef unsigned long long ull;
const int maxn = 1e6 + 10;
const int  maxm = 300;
const int turn[4][2] = {{1, 0}, { -1, 0}, {0, 1}, {0, -1}};
//next_permutation
int n, now, cur, pop;
ll mod = 1e9 + 7;
int a[maxn];
int visit[maxn];
int flag = 0;
int gcd(int x, int y)
{
        if (y == 0)
        {
                return x;
        }
        else
        {
                return (gcd(y, x % y));
        }
}
int main()
{
        cin >> n;
        for (int i = 1; i <= n; i++)
        {
                scanf("%d", &a[i]);
                visit[a[i]] = 1;
        }
        int now = 0;
        for (int i = 1; i <= n; ++i)
        {
                now = gcd(now, a[i]);
        }
        if (now != a[1])
        {
                cout << -1 << endl;
                return 0;
        }
        else
        {
                cout << n * 2 << endl;
                for (int i = 1; i <= n; ++i)
                {
                        cout << a[i] << " " << a[1] << " ";
                }
                cout << endl;
        }
        return 0;
}

 D:

#include <iostream>
#include <stdio.h>
#include <math.h>
#include <string.h>
#include <time.h>
#include <stdlib.h>
#include <string>
#include <bitset>
#include <vector>
#include <set>
#include <map>
#include <queue>
#include <algorithm>
#include <sstream>
#include <stack>
#include <iomanip>
using namespace std;
#define pb push_back
#define mp make_pair
typedef pair<int,int> pii;
typedef long long ll;
typedef double ld;
typedef vector<int> vi;
#define fi first
#define se second
#define fe first
#define FO(x) {freopen(#x".in","r",stdin);freopen(#x".out","w",stdout);}
#define Edg int M=0,fst[SZ],vb[SZ],nxt[SZ];void ad_de(int a,int b){++M;nxt[M]=fst[a];fst[a]=M;vb[M]=b;}void adde(int a,int b){ad_de(a,b);ad_de(b,a);}
#define Edgc int M=0,fst[SZ],vb[SZ],nxt[SZ],vc[SZ];void ad_de(int a,int b,int c){++M;nxt[M]=fst[a];fst[a]=M;vb[M]=b;vc[M]=c;}void adde(int a,int b,int c){ad_de(a,b,c);ad_de(b,a,c);}
#define es(x,e) (int e=fst[x];e;e=nxt[e])
#define esb(x,e,b) (int e=fst[x],b=vb[e];e;e=nxt[e],b=vb[e])
#define SZ 2333333
int n,m,l[SZ];
vector<int> vs[1000099];
ll ans[SZ];
vector<pii> qs[1000099];
inline void add_qry(int a,int b,int c)
{
    if(c>=0) qs[b].pb(pii(a,c));
}
int ts[SZ]; ll qzh[SZ];
#define UP 23333333
void dfs(int x)
{
    if(x+x<=n)
    {
        dfs(x+x);
        for(auto g:vs[x+x])
            if(g+l[x+x]<=UP) vs[x].pb(g+l[x+x]);
        vs[x+x].clear();
    }
    if(x+x+1<=n)
    {
        dfs(x+x+1);
        for(auto g:vs[x+x+1])
            if(g+l[x+x+1]<=UP) vs[x].pb(g+l[x+x+1]);
        vs[x+x+1].clear();
    }
    vs[x].pb(0);
    int tn=0;
    for(auto g:vs[x]) ts[++tn]=g;
    sort(ts+1,ts+1+tn);
    for(int i=1;i<=tn;++i)
        qzh[i]=qzh[i-1]+ts[i];
    for(auto q:qs[x])
    {
        int c=q.se,u=upper_bound(ts+1,ts+1+tn,c)-ts-1;
        ans[q.fi]+=c*(ll)u-qzh[u];
    }
}
int main()
{
    scanf("%d%d",&n,&m);
    for(int i=2;i<=n;++i) scanf("%d",l+i);
    for(int i=1;i<=m;++i)
    {
        int a,b,bf=0;
        scanf("%d%d",&a,&b);
        add_qry(i,a,b);
        for(int g=a;g&&b>=0;bf=g,b-=l[g],g>>=1)
        {
            if(g==a) continue;
            ans[i]+=b;
            add_qry(i,bf^1,b-l[bf^1]);
        }
    }
    dfs(1);
    for(int i=1;i<=m;++i)
        printf("%I64d\n",ans[i]);
}