每日三题

9.27

A - Moore's Law

思路：快速幂

#include<bits/stdc++.h>
using namespace std;
#define int long long
//#define int __int128
#define double long double
typedef pair<int,int>PII;
typedef pair<string,int>PSI;
typedef pair<string,string>PSS;
const int N=1e3+5,INF=0x3f3f3f3f,Mod=1e9+7,mod=998244353;
const double eps=1e-6;

double qsm(double a,int b){
    double res=(double)1;
    while(b){
        if(b&1)res*=a;
        b>>=1;
        a*=a;
    }
    return res;
}
void solve(){
    int n,t;cin>>n>>t;
    double ans,p=1.000000011;
    ans=(double)n*qsm(p,t);
    cout<<fixed<<setprecision(20)<<ans;
}
signed main(){
    ios::sync_with_stdio(0),cin.tie(0),cout.tie(0);
    int T=1;
    //cin>>T;
    while(T--){
        solve();
    }
    return 0;
}
View Code

 

B - Load Balancing

思路：若sum能整除n，那么把所有数变为sum/n；否则，有sum%n个数变为sum/n+1，那么将大于sum/n+1的数变为sum/n+1即可

#include<bits/stdc++.h>
using namespace std;
#define int long long
//#define int __int128
#define double long double
typedef pair<int,int>PII;
typedef pair<string,int>PSI;
typedef pair<string,string>PSS;
const int N=1e3+5,INF=0x3f3f3f3f,Mod=1e9+7,mod=998244353;
const double eps=1e-6;


void solve(){
    int s=0,n;cin>>n;
    vector<int>ve(n);
    for(int i=0;i<n;++i){
        cin>>ve[i];s+=ve[i];
    }
    int ans=0,b=0,a=s/n;
    if(s%n){
        b=s%n;
        sort(ve.begin(),ve.end());
        for(int i=n-1;i>=0;--i){
            if(b){
                ans+=abs(ve[i]-a-1);b--;
            }else ans+=abs(ve[i]-a);
        }
    }
    else{
        for(int i=0;i<n;++i){
            ans+=abs(ve[i]-a);
        }
    }
    cout<<ans/2;
}
signed main(){
    ios::sync_with_stdio(0),cin.tie(0),cout.tie(0);
    int T=1;
    //cin>>T;
    while(T--){
        solve();
    }
    return 0;
}
View Code

 

C - The Union of k-Segments

思路：可以把每个区间的端点的位置以及是左/右端点 统计起来，按位置前、左端点在前排序；sum统计经过的端点数，经过左端点sum++，右端点sum--，当sum=k时，无论是经过左还是右端点，都是一段有效区间答案的端点，都统计到答案里

#include<bits/stdc++.h>
using namespace std;
#define int long long
//#define int __int128
#define double long double
typedef pair<int,int>PII;
typedef pair<string,int>PSI;
typedef pair<string,string>PSS;
const int N=1e3+5,INF=0x3f3f3f3f,Mod=1e9+7,mod=998244353;
const double eps=1e-6;

struct E{
    int st,op;//10
    bool operator<(const E&e)const{
        if(st==e.st)return op>e.op;
        return st<e.st;
    }
};
void solve(){
    int k,n;cin>>n>>k;
    vector<E>ve(2*n);
    for(int i=0,p=0;i<n;++i){
        cin>>ve[p].st>>ve[p+1].st;
        ve[p++].op=1,ve[p++].op=0;
    }
    sort(ve.begin(),ve.end());
    vector<int>ans;
    for(int i=0,s=0;i<2*n;++i){
        if(ve[i].op){
            s++;
            if(s==k)ans.push_back(ve[i].st);
        }else{
            if(s==k)ans.push_back(ve[i].st);
            s--;
        }
    }
    cout<<ans.size()/2<<'\n';
    for(int i=0;i<ans.size();i+=2){
        cout<<ans[i]<<' '<<ans[i+1]<<'\n';
    }
}
signed main(){
    ios::sync_with_stdio(0),cin.tie(0),cout.tie(0);
    int T=1;
    //cin>>T;
    while(T--){
        solve();
    }
    return 0;
}
View Code

 

9.28

A - New Skateboard

思路：由于是找满足条件的子串，对于最后一位是0、4、8的数，只要前面组成的数为偶数，即前一位为偶数，一定可以与前面所有的数组成满足条件的数；对于最后一位是2、6的数，只要前面组成的数为奇数，即前一位为奇数，一定可以与前面所有数组成满足条件的数。

#include<bits/stdc++.h>
using namespace std;
#define int long long
//#define int __int128
#define double long double
typedef pair<int,int>PII;
typedef pair<string,int>PSI;
typedef pair<string,string>PSS;
const int N=1e3+5,INF=0x3f3f3f3f,Mod=1e9+7,mod=998244353;
const double eps=1e-6;


void solve(){
    string s;cin>>s;
    int n=s.size(),ans;
    s=" "+s;
    vector<int>f(n+1);
    if((s[1]-'0')%4==0)f[1]=1;
    ans=f[1];
    for(int i=2;i<=n;++i){
        if(s[i]=='0'||s[i]=='4'||s[i]=='8'){
            if((s[i-1]-'0')%2==0)f[i]+=i-1;
            f[i]++;
        }else if(s[i]=='2'||s[i]=='6'){
            if((s[i-1]-'0')%2)f[i]+=i-1;
        }
        ans+=f[i];
    }
    cout<<ans;
}
signed main(){
    ios::sync_with_stdio(0),cin.tie(0),cout.tie(0);
    int T=1;
    //cin>>T;
    while(T--){
        solve();
    }
    return 0;
}
View Code

 

B - Replace To Make Regular Bracket Sequence

思路：用栈存左边，遇到右边判断下是否相同即可

#include<bits/stdc++.h>
using namespace std;
#define int long long
//#define int __int128
#define double long double
typedef pair<int,int>PII;
typedef pair<string,int>PSI;
typedef pair<string,string>PSS;
const int N=1e3+5,INF=0x3f3f3f3f,Mod=1e9+7,mod=998244353;
const double eps=1e-6;


void solve(){
    string s;cin>>s;
    stack<char>st;
    map<char,int>mp;
    mp['(']=mp[')']=1;
    mp['[']=mp[']']=2;
    mp['{']=mp['}']=3;
    mp['<']=mp['>']=4;
    int ans=0;
    for(int i=0;i<s.size();++i){
        if(s[i]=='<'||s[i]=='('||s[i]=='{'||s[i]=='['){
            st.push(s[i]);
        }else{
            if(st.empty()){
                cout<<"Impossible";return ;
            }
            if(mp[st.top()]!=mp[s[i]]){
                ans++;
            }
            st.pop();
        }
    }
    if(st.empty())cout<<ans;
    else cout<<"Impossible";
}
signed main(){
    ios::sync_with_stdio(0),cin.tie(0),cout.tie(0);
    int T=1;
    //cin>>T;
    while(T--){
        solve();
    }
    return 0;
}
View Code

 

C - The Labyrinth

思路：dfs把联通的'.'个数求出来，并给这个连通块编号。每个'*'遍历四个方向的'.'连通块即可（答案要%10！）

#include<bits/stdc++.h>
using namespace std;
#define int long long
//#define int __int128
#define double long double
typedef pair<int,int>PII;
typedef pair<string,int>PSI;
typedef pair<string,string>PSS;
const int N=1e3+5,INF=0x3f3f3f3f,Mod=1e9+7,mod=998244353;
const double eps=1e-6;
string s[N];
int st[N][N],sum,vis[N*N];
int ans[N][N],n,m,idx,cnt[N*N];
int dx[4]={-1,0,1,0};
int dy[4]={0,1,0,-1};
void dfs(int x,int y){
    for(int i=0;i<4;++i){
        int xx=x+dx[i],yy=y+dy[i];
        if(xx>=1&&xx<=n&&yy>=1&&yy<=m&&st[xx][yy]==0&&s[xx][yy]=='.'){
            st[xx][yy]=idx;
            sum++;
            dfs(xx,yy);
        }
    }
}
void solve(){
    cin>>n>>m;
    for(int i=1;i<=n;++i){
        cin>>s[i];
        s[i]=" "+s[i];
    }
    for(int i=1;i<=n;++i){
        for(int j=1;j<=m;++j){
            if(st[i][j]==0&&s[i][j]=='.'){
                sum=1;
                st[i][j]=++idx;
                dfs(i,j);
                cnt[idx]=sum;
            }
        }
    }
    for(int i=1;i<=n;++i){
        for(int j=1;j<=m;++j){
            if(s[i][j]=='.')cout<<'.';
            else {
                int c=1;
                for(int k=0;k<4;++k){
                    int x=i+dx[k],y=j+dy[k];
                    if(x>=1&&x<=n&&y>=1&&y<=m&&vis[st[x][y]]!=(n*(i-1)+j)){
                        vis[st[x][y]]=(n*(i-1)+j);
                        c+=cnt[st[x][y]];
                    }
                }
                cout<<c%10;
            }
        }cout<<'\n';
    }
}
signed main(){
    ios::sync_with_stdio(0),cin.tie(0),cout.tie(0);
    int T=1;
    //cin>>T;
    while(T--){
        solve();
    }
    return 0;
}
View Code

 9.29

A - Average Sleep Time

思路：取每个长度为k的连续段的和，求平均值

#include<bits/stdc++.h>
using namespace std;
#define int long long
#define PII pair<int,int>
const int mod=1e9+7;
void solve(){
    int n,k;cin>>n>>k;
    int s=0;
    vector<int>ve(n+1);
    for(int i=1;i<=n;++i){
        cin>>ve[i];ve[i]+=ve[i-1];
    }
    for(int i=0;i<=n-k;++i){
        s+=ve[i+k]-ve[i];
    }
    double ans=(double)s/(n-k+1);
    cout<<fixed<<setprecision(9)<<ans<<'\n';
}
signed main(){
    ios::sync_with_stdio(false),cin.tie(0),cout.tie(0);
    int _=1;
    //cin>>_;
    while (_--)
        solve();
}
View Code

 

B - Gabriel and Caterpillar

思路：第一次的h1、h2为第一次下课见到的高度，白天+黑夜算做一天，由于从白天开始，且白天是上升，那么第一次晚上后算作第一天开始，计算出一天内上升的高度，求需要多少天即可；

#include<bits/stdc++.h>
using namespace std;
#define int long long
//#define int __int128
#define double long double
typedef pair<int,int>PII;
typedef pair<string,int>PSI;
typedef pair<string,string>PSS;
const int N=1e3+5,INF=0x3f3f3f3f,Mod=1e9+7,mod=998244353;
const double eps=1e-6;

void solve(){
    int h1,h2,a,b;cin>>h1>>h2>>a>>b;
    h1+=8*a;
    if(h1>=h2)cout<<0;
    else{
        int c=h2-h1,d=12*(a-b);
        if(d<=0)cout<<-1;
        else cout<<(c+d-1)/d;
    }
}
signed main(){
    ios::sync_with_stdio(0),cin.tie(0),cout.tie(0);
    int T=1;
    //cin>>T;
    while(T--){
        solve();
    }
    return 0;
}
View Code

 

C - Longest k-Good Segment

思路：用双指针维护一下即可

#include<bits/stdc++.h>
using namespace std;
#define int long long
//#define int __int128
#define double long double
typedef pair<int,int>PII;
typedef pair<string,int>PSI;
typedef pair<string,string>PSS;
const int N=1e3+5,INF=0x3f3f3f3f,Mod=1e9+7,mod=998244353;
const double eps=1e-6;

void solve(){
    int n,k;cin>>n>>k;
    vector<int>ve(n+1);
    for(int i=0;i<n;++i)cin>>ve[i];
    int ma=0,l,r;
    map<int,int>mp;
    for(int i=0,j=0;i<n;++i){
        if(mp.size()<k)mp[ve[i]]++;
        else{
            if(mp[ve[i]]==0){
                mp[ve[j]]--;
                while(mp[ve[j]]!=0&&j+1<=i) {
                    mp[ve[++j]]--;
                }
                j++;
            }mp[ve[i]]++;
        }
        if(i-j+1>ma){
            ma=i-j+1;
            l=j,r=i;
        }
    }
    cout<<l+1<<' '<<r+1;
}
signed main(){
    ios::sync_with_stdio(0),cin.tie(0),cout.tie(0);
    int T=1;
    //cin>>T;
    while(T--){
        solve();
    }
    return 0;
}
View Code

 

9.30

A - Distances to Zero

思路：维护前后最近的0的位置即可

#include<bits/stdc++.h>
using namespace std;
#define int long long
//#define int __int128
#define double long double
typedef pair<int,int>PII;
typedef pair<string,int>PSI;
typedef pair<string,string>PSS;
const int N=1e3+5,INF=0x3f3f3f3f,Mod=1e9+7,mod=998244353;
const double eps=1e-6;

void solve(){
    int n;cin>>n;
    vector<int>ve(n+1),l(n+5),r(n+5);
    for(int i=1,last=0;i<=n;++i){
        cin>>ve[i];
        if(ve[i]==0)last=i;
        else{
            if(last)l[i]=i-last;
            else l[i]=INF;
        }
    }
    for(int i=n,last=0;i>=1;--i){
        if(ve[i]==0)last=i;
        else{
            if(last)r[i]=last-i;
            else r[i]=INF;
        }
    }
    for(int i=1;i<=n;++i){
        cout<<min(l[i],r[i])<<' ';
    }

}
signed main(){
    ios::sync_with_stdio(0),cin.tie(0),cout.tie(0);
    int T=1;
    //cin>>T;
    while(T--){
        solve();
    }
    return 0;
}
View Code

 

B - Shopping

思路：模拟下过程即可（题目给的是每个位置的物品，每次加的是该物品的位置）

#include<bits/stdc++.h>
using namespace std;
#define int long long
//#define int __int128
#define double long double
typedef pair<int,int>PII;
typedef pair<string,int>PSI;
typedef pair<string,string>PSS;
const int N=1e3+5,INF=0x3f3f3f3f,Mod=1e9+7,mod=998244353;
const double eps=1e-6;

void solve(){
    int n,k,m,ans=0;cin>>n>>m>>k;
    vector<int>ve(k+1),pos(k+1);
    for(int i=1;i<=k;++i)cin>>ve[i],pos[ve[i]]=i;
    for(int i=1;i<=n;++i){
        for(int j=1,x;j<=m;++j){
            cin>>x;
            int p=pos[x];
            ans+=p;
            for(int v=p;v>1;--v){
                ve[v]=ve[v-1];
                pos[ve[v]]=v;
            }
            ve[1]=x;
            pos[x]=1;
        }
    }
    cout<<ans;
}
signed main(){
    ios::sync_with_stdio(0),cin.tie(0),cout.tie(0);
    int T=1;
    //cin>>T;
    while(T--){
        solve();
    }
    return 0;
}
View Code

 

C - Not Equal on a Segment

思路：可以维护前缀pre，表示前面第一个不同的数的位置；也可以用数据结构维护区间最大和最小值及位置；

#include<bits/stdc++.h>
using namespace std;
#define int long long
//#define int __int128
#define double long double
typedef pair<int,int>PII;
typedef pair<string,int>PSI;
typedef pair<string,string>PSS;
const int N=1e6+5,INF=0x3f3f3f3f,Mod=1e9+7,mod=998244353;
const double eps=1e-6;

void solve(){
    int n,m;cin>>n>>m;
    vector<int>ve(n+1),pre(n+1);
    for(int i=1;i<=n;++i){
        cin>>ve[i];
        if(i==1)pre[i]=0;
        else{
            if(ve[i]!=ve[i-1])pre[i]=i-1;
            else pre[i]=pre[i-1];
        }
    }
    for(int i=1,l,r,x;i<=m;++i){
        cin>>l>>r>>x;
        if(ve[r]!=x)cout<<r<<'\n';
        else if(pre[r]<l)cout<<-1<<'\n';
        else cout<<pre[r]<<'\n';
    }
}
signed main(){
    ios::sync_with_stdio(0),cin.tie(0),cout.tie(0);
    int T=1;
    //cin>>T;
    while(T--){
        solve();
    }
    return 0;
}
View Code

 

10.1

A - Treasure Hunt

思路：判断x，y轴转换需要的次数是否同奇或同偶

#include<bits/stdc++.h>
using namespace std;
#define int long long
//#define int __int128
#define double long double
typedef pair<int,int>PII;
typedef pair<string,int>PSI;
typedef pair<string,string>PSS;
const int N=1e6+5,INF=0x3f3f3f3f,Mod=1e9+7,mod=998244353;
const double eps=1e-6;

void solve(){
    int x1,y1,x2,y2,x,y;
    cin>>x1>>y1>>x2>>y2>>x>>y;
    int c1=abs(x1-x2),c2=abs(y1-y2);
    if(c1%x||c2%y)cout<<"NO";
    else{
        c1/=x,c2/=y;
        if(c1%2==c2%2)cout<<"YES";
        else cout<<"NO";
    }
}
signed main(){
    ios::sync_with_stdio(0),cin.tie(0),cout.tie(0);
    int T=1;
    //cin>>T;
    while(T--){
        solve();
    }
    return 0;
}
View Code

 

 B - Optimal Point on a Line

思路：定为最中间的点最优

#include<bits/stdc++.h>
using namespace std;
#define int long long
//#define int __int128
#define double long double
typedef pair<int,int>PII;
typedef pair<string,int>PSI;
typedef pair<string,string>PSS;
const int N=1e6+5,INF=0x3f3f3f3f,Mod=1e9+7,mod=998244353;
const double eps=1e-6;

void solve(){
    int n;cin>>n;
    vector<int>ve(n+1);
    for(int i=1;i<=n;++i)cin>>ve[i];
    sort(ve.begin()+1,ve.end());
    if(n%2)cout<<ve[n/2+1];
    else cout<<ve[n/2];
}
signed main(){
    ios::sync_with_stdio(0),cin.tie(0),cout.tie(0);
    int T=1;
    //cin>>T;
    while(T--){
        solve();
    }
    return 0;
}
View Code

 

 C - Foe Pairs

思路：dp，维护以i结尾有多少个合法区间，维护下i前的第一个pair的左端点即可知道以i结尾的合法区间有多少个，那么记录每对的右端点的最近的左端点

#include<bits/stdc++.h>
using namespace std;
#define int long long
//#define int __int128
#define double long double
typedef pair<int,int>PII;
typedef pair<string,int>PSI;
typedef pair<string,string>PSS;
const int N=1e6+5,INF=0x3f3f3f3f,Mod=1e9+7,mod=998244353;
const double eps=1e-6;

void solve(){
    int n,m;cin>>n>>m;
    vector<int>ve(n+1),pos(n+1),las(n+1);
    for(int i=1;i<=n;++i){
        cin>>ve[i];
        pos[ve[i]]=i;
    }
    for(int i=1,x,y;i<=m;++i){
        cin>>x>>y;
        if(pos[x]>pos[y])swap(x,y);
        las[y]=max(las[y],pos[x]);
    }
    int ans=0,pre=0;
    for(int i=1;i<=n;++i){
        if(las[ve[i]])pre=max(pre,las[ve[i]]);
        ans+=i-pre;
    }
    cout<<ans;
}
signed main(){
    ios::sync_with_stdio(0),cin.tie(0),cout.tie(0);
    int T=1;
    //cin>>T;
    while(T--){
        solve();
    }
    return 0;
}
View Code

 

10.2

A - Counting-out Rhyme

思路：模拟每一步即可

#include<bits/stdc++.h>
using namespace std;
#define int long long
//#define int __int128
#define double long double
typedef pair<int,int>PII;
typedef pair<string,int>PSI;
typedef pair<string,string>PSS;
const int N=1e6+5,INF=0x3f3f3f3f,Mod=1e9+7,mod=998244353;
const double eps=1e-6;

void solve(){
    int st=1,n,k;cin>>n>>k;
    vector<int>ve(k+1),f(n+1);
    for(int i=1;i<=n;++i)f[i]=i;
    for(int i=1;i<=k;++i)cin>>ve[i];
    for(int i=1;i<=k;++i){
        int p=ve[i]+st;
        int now=p%n;
        if(now==0)now=n;
        cout<<f[now]<<' ';
        n--;
        if(now!=n+1)
        for(int j=now;j<=n;++j)f[j]=f[j+1];
        else now=1;
        st=now;
//        for(int j=1;j<=n;++j)cout<<f[j]<<' ';cout<<'\n';
    }
}
signed main(){
    ios::sync_with_stdio(0),cin.tie(0),cout.tie(0);
    int T=1;
    //cin>>T;
    while(T--){
        solve();
    }
    return 0;
}
View Code

 

B - Magic Odd Square

 思路：由于n为奇数，且要满足条件每一行/列/对角线的和为奇数，则每一行/列/对角线的奇数个数为奇数个，且总奇数数有一半，那么从第一行起每一行奇数个数为1、3、5...，在中心且对称，如下图

00100
01110
11111
01110
00100

 

#include<bits/stdc++.h>
using namespace std;
#define int long long
//#define int __int128
#define double long double
typedef pair<int,int>PII;
typedef pair<string,int>PSI;
typedef pair<string,string>PSS;
const int N=1e6+5,INF=0x3f3f3f3f,Mod=1e9+7,mod=998244353;
const double eps=1e-6;

void solve(){
    int n,ou=2,ji=1;cin>>n;
    int m=(n+1)/2;
    for(int i=1,l,r;i<=n;++i){
        if(i<=m)l=m-i+1,r=m+i-1;
        else l=i-m+1,r=n-i+m;
        for(int j=1;j<=n;++j){
            if(j>=l&&j<=r)cout<<ji,ji+=2;
            else cout<<ou,ou+=2;
            cout<<' ';
        }cout<<'\n';
    }
}
signed main(){
    ios::sync_with_stdio(0),cin.tie(0),cout.tie(0);
    int T=1;
    //cin>>T;
    while(T--){
        solve();
    }
    return 0;
}
View Code

 

C - Hard Process

思路：可以求以i结尾的最长子串，那么就要求出以i结尾最多能换的0的个数，可以二分求，用前缀和判断，也可以用队列预处理出来。

#include<bits/stdc++.h>
using namespace std;
#define int long long
//#define int __int128
#define double long double
typedef pair<int,int>PII;
typedef pair<string,int>PSI;
typedef pair<string,string>PSS;
const int N=1e6+5,INF=0x3f3f3f3f,Mod=1e9+7,mod=998244353;
const double eps=1e-6;

void solve(){
    int n,k;cin>>n>>k;
    vector<int>ve(n+1),f(n+1),las(n+1);
    queue<int>q;
    for(int i=1,l=0;i<=n;++i){
        cin>>ve[i];
        las[i]=l;
        if(ve[i]==0)q.push(i),l=i;
        if(q.size()>k)q.pop();
        if(q.size())f[i]=q.front();
    }
//    for(int i=1;i<=n;++i)cout<<f[i]<<' ';cout<<'\n';
    int ma=0,ansl,ansr;
    for(int i=1,l;i<=n;++i){
        int c=0;
        if(f[i])c=i-las[f[i]];
        else if(ve[i]) c=i-las[i];
        if(c>ma){
            ma=c;
            ansl=i-c+1,ansr=i;
        }//cout<<c<<' ';
    }
    cout<<ma<<'\n';
    for(int i=1;i<=n;++i){
        if(i>=ansl&&i<=ansr)cout<<1<<' ';
        else cout<<ve[i]<<' ';
    }
}
signed main(){
    ios::sync_with_stdio(0),cin.tie(0),cout.tie(0);
    int T=1;
    //cin>>T;
    while(T--){
        solve();
    }
    return 0;
}
View Code

 

10.3

A - Simple Strings

思路：对于连续相同的一段字母，若长度为奇数，从第二个开始每隔一个改变；若为偶数，从第一个开始每隔一个改变，可以保证这样的次数最少

#include<bits/stdc++.h>
using namespace std;
#define int long long
//#define int __int128
#define double long double
typedef pair<int,int>PII;
typedef pair<string,int>PSI;
typedef pair<string,string>PSS;
const int N=5e5+5,INF=0x3f3f3f3f,Mod=1e9+7,mod=998244353;
const double eps=1e-6;


void solve(){
    string s;cin>>s;
    for(int i=0,r,j;i<s.size();++i){
        r=i;
        while(r+1<s.size()&&s[r+1]==s[i])r++;
        char c=s[i];
        while(c==s[i]||(r<s.size()-1&&c==s[r+1])||(i&&s[i-1]==c)){
            if(c=='z')c='a';
            else c++;
        }
        if((r-i+1)%2)j=i+1;
        else j=i;
        for(;j<=r;j+=2)s[j]=c;
        i=r;
    }
    cout<<s;
}

signed main(){
    ios::sync_with_stdio(0),cin.tie(0),cout.tie(0);
    int t=1;
    //cin>>t;
    while(t--){
        solve();
    }
    return 0;
}
View Code

 

B - Odd sum

思路：奇+奇=偶，奇+偶=奇；对于所有整偶数直接加即可，将所有整奇数加了之后，若奇数个数为偶数，需要减去一个最小的正奇数或加上一个最大的负奇数

#include<bits/stdc++.h>
using namespace std;
#define int long long
//#define int __int128
#define double long double
typedef pair<int,int>PII;
typedef pair<string,int>PSI;
typedef pair<string,string>PSS;
const int N=5e5+5,INF=0x3f3f3f3f,Mod=1e9+7,mod=998244353;
const double eps=1e-6;


void solve(){
    int n;cin>>n;
    int ans=0,cnt1=0,mi=INF,ma=-INF;
    for(int x,i=0;i<n;++i){
        cin>>x;
        if(x>=0){
            if(x%2)ans+=x,cnt1++,mi=min(mi,x);
            else ans+=x;
        }else if(x%2)ma=max(ma,x);
    }
    if(cnt1%2==0)ans-=min(mi,abs(ma));
    cout<<ans;
}

signed main(){
    ios::sync_with_stdio(0),cin.tie(0),cout.tie(0);
    int t=1;
    //cin>>t;
    while(t--){
        solve();
    }
    return 0;
}
View Code

 

C - Nested Segments

思路：若区间i包含区间j，需满足li<=lj和rj<=ri；先按左端点排序（右端点也可）后满足了一个条件；对于第i个区间，第i个后的所有区间中满足rj<=ri的区间个数为第i区间包含的区间，可以用树状数组单点加及区间和来求；由于r范围较大，离散化后便可用树状数组求

#include<bits/stdc++.h>
using namespace std;
#define int long long
//#define int __int128
#define double long double
typedef pair<int,int>PII;
typedef pair<string,int>PSI;
typedef pair<string,string>PSS;
const int N=2e5+5,INF=0x3f3f3f3f,Mod=1e9+7,mod=998244353;
const double eps=1e-6;
struct E{
    int l,r,id;
};
int n;
vector<int>f(N);
int lowbit(int x){return x&-x;}
int getsum(int x){
    int ans=0;
    while(x>0){
        ans+=f[x];
        x-=lowbit(x);
    }
    return ans;
}
void add(int x,int k){
    while(x<=n){
        f[x]+=k;
        x+=lowbit(x);
    }
}
bool cmp1(E a,E b){return a.l<b.l;}
bool cmp2(E a,E b){return a.r<b.r;}

void solve(){
    cin>>n;
    vector<E>ve(n+1);
    for(int i=1;i<=n;++i)cin>>ve[i].l>>ve[i].r,ve[i].id=i;
    sort(ve.begin()+1,ve.end(),cmp1);
    for(int i=1;i<=n;++i)ve[i].l=i;
    sort(ve.begin()+1,ve.end(),cmp2);
    vector<int>ans(n+1);
    for(int i=1;i<=n;++i){
        ans[ve[i].id]=i-1- getsum(ve[i].l-1);
        add(ve[i].l,1);
    }
    for(int i=1;i<=n;++i)cout<<ans[i]<<'\n';
}
signed main(){
    ios::sync_with_stdio(0),cin.tie(0),cout.tie(0);
    int T=1;
    //cin>>T;
    while(T--){
        solve();
    }
    return 0;
}
View Code

 

10.4

A - Co-prime Array

思路：1和任何正整数都互质，若存在不满足的插入1即可

#include<bits/stdc++.h>
using namespace std;
#define int long long
//#define int __int128
#define double long double
typedef pair<int,int>PII;
typedef pair<string,int>PSI;
typedef pair<string,string>PSS;
const int N=5e5+5,INF=0x3f3f3f3f,Mod=1e9+7,mod=998244353;
const double eps=1e-6;


void solve(){
    int n;cin>>n;
    vector<int>ve(n),ans;
    for(int i=0;i<n;++i){
        cin>>ve[i];
    }
    for(int i=0;i<n-1;++i){
        ans.push_back(ve[i]);
        if(__gcd(ve[i],ve[i+1])!=1)ans.push_back(1);
    }ans.push_back(ve[n-1]);
    cout<<ans.size()-n<<'\n';
    for(auto v:ans)cout<<v<<' ';
}
signed main(){
    ios::sync_with_stdio(0),cin.tie(0),cout.tie(0);
    int t=1;
    //cin>>t;
    while(t--){
        solve();
    }
    return 0;
}
View Code

 

B - Tea Party

思路：倒的茶要单调，先倒一半，剩余的从高开始补满，还剩多的或不够倒即不存在

#include<bits/stdc++.h>
using namespace std;
#define int long long
//#define int __int128
#define double long double
typedef pair<int,int>PII;
typedef pair<string,int>PSI;
typedef pair<string,string>PSS;
const int N=5e5+5,INF=0x3f3f3f3f,Mod=1e9+7,mod=998244353;
const double eps=1e-6;


void solve(){
    int n,w,s=0;cin>>n>>w;
    vector<PII>a(n+1);
    for(int i=1;i<=n;++i)cin>>a[i].first,a[i].second=i,s+=(a[i].first+1)/2;
    if(s>w){
        cout<<-1;return ;
    }
    sort(a.begin()+1,a.end());
    int l=w-s;
    vector<int>ans(n+1);
    for(int i=n;i&&l;--i){
        int c=min(a[i].first/2,l);
        l-=c;
        ans[a[i].second]=(a[i].first+1)/2+c;
    }
    if(l){
        cout<<-1;return ;
    }
    for(int i=1;i<=n;++i){
        if(ans[a[i].second]==0)ans[a[i].second]=(a[i].first+1)/2;
    }
    for(int i=1;i<=n;++i)cout<<ans[i]<<' ';
}
signed main(){
    ios::sync_with_stdio(0),cin.tie(0),cout.tie(0);
    int t=1;
    //cin>>t;
    while(t--){
        solve();
    }
    return 0;
}
View Code

 

C - Simple Subset

思路：由于奇+奇=偶+偶=偶，那么数组里最多有一个奇数和一个偶数，特别的，1可以存在多个（1+1=2，2为素数）；

那么就有两种情况：多个1和一个偶，一个奇和一个偶；取长度最多的情况

若找不到和为素数的情况，则任意一个数都可以

#include<bits/stdc++.h>
using namespace std;
#define int long long
//#define int __int128
#define double long double
typedef pair<int,int>PII;
typedef pair<string,int>PSI;
typedef pair<string,string>PSS;
const int N=2e6+5,INF=0x3f3f3f3f,Mod=1e9+7,mod=998244353;
const double eps=1e-6;
int st[N],primes[N],idx;
void init(){
    for(int i=2;i<N;++i){
        if(!st[i])
            primes[idx++]=i;
        for(int j=0;primes[j]*i<N;++j){
            st[primes[j]*i]=1;
            if(i%primes[j]==0)break;
        }
    }
}
void solve(){
    int n;cin>>n;
    int cnt1=0;
    vector<int>ve(n),ji,ou;
    int ok1=0;
    for(int i=0;i<n;++i){
        cin>>ve[i];
        if(ve[i]%2)ji.push_back(ve[i]);
        else {
            ou.push_back(ve[i]);
            if(!st[ve[i]+1])ok1=ve[i];
        }
        if(ve[i]==1)cnt1++;
    }
    vector<int>ans;
    if(cnt1){
        for(int i=1;i<=cnt1;++i)ans.push_back(1);
        if(ok1)ans.push_back(ok1);
    }
    if(ans.size()<=1){
        for(int i=0;i<ji.size();++i){
            for(int j=0;j<ou.size();++j){
                if(!st[ji[i]+ou[j]]){
                    ans.clear();
                    ans.push_back(ji[i]),ans.push_back(ou[j]);
                    break;
                }
            }
        }
    }
    if(ans.size()==0)ans.push_back(ve[0]);
    cout<<ans.size()<<'\n';
    for(auto v:ans)cout<<v<<' ';
}
signed main(){
    ios::sync_with_stdio(0),cin.tie(0),cout.tie(0);
    int T=1;
    //cin>>T;
    init();
    while(T--){
        solve();
    }
    return 0;
}
View Code

 

10.5

A - Grandma Laura and Apples

思路：由于最后剩0个，且每次买一半，倒推回去即可

#include<bits/stdc++.h>
using namespace std;
#define int long long
//#define int __int128
#define double long double
typedef pair<int,int>PII;
typedef pair<string,int>PSI;
typedef pair<string,string>PSS;
const int N=2e6+5,INF=0x3f3f3f3f,Mod=1e9+7,mod=998244353;
const double eps=1e-6;

void solve(){
    int n,p;cin>>n>>p;
    int ans=0;
    vector<string>ve(n);
    for(int i=0;i<n;++i)cin>>ve[i];
    for(int s=0,i=n-1;i>=0;--i){
        if(ve[i]=="halfplus")s=s*2+1;
        else s*=2;
        ans+=s*10/2;
    }
    ans=ans/10*p+p/2*(ans%10!=0);
    cout<<ans;
}
signed main(){
    ios::sync_with_stdio(0),cin.tie(0),cout.tie(0);
    int T=1;
    //cin>>T;
//    init();
    while(T--){
        solve();
    }
    return 0;
}
View Code

 

B - Suitable Replacement

思路：记录下t串每个字符的个数，对于'?'按顺序凑满每个t即可

#include<bits/stdc++.h>
using namespace std;
#define int long long
//#define int __int128
#define double long double
typedef pair<int,int>PII;
typedef pair<string,int>PSI;
typedef pair<string,string>PSS;
const int N=2e6+5,INF=0x3f3f3f3f,Mod=1e9+7,mod=998244353;
const double eps=1e-6;

void solve(){
    string s,t;cin>>s>>t;
    map<char,int>mp,f;
    for(int i=0;i<t.size();++i)mp[t[i]]++;
    for(int i=0;i<s.size();++i){
        if(s[i]!='?'&&mp[s[i]]>0)f[s[i]]++;
    }
    for(int i=0,j=0;i<s.size();++i){
        if(s[i]=='?'){
            while(f[t[j]]>=mp[t[j]]){
                if(j+1==t.size()){
                    j=0;
                    for(auto &v:f){
                        v.second=max(0ll,v.second-mp[v.first]);
                    }
                }else j++;
            }
            f[t[j]]++;
            s[i]=t[j];
        }
    }
    cout<<s;
}
signed main(){
    ios::sync_with_stdio(0),cin.tie(0),cout.tie(0);
    int T=1;
    //cin>>T;
//    init();
    while(T--){
        solve();
    }
    return 0;
}
View Code

 

C - Joty and Chocolate

思路：涂红色的是a的倍数，蓝色是b的倍数，对与lcm（a，b）的倍数两个颜色都能涂，那么涂价值最高的即可

#include<bits/stdc++.h>
using namespace std;
#define int long long
//#define int __int128
#define double long double
typedef pair<int,int>PII;
typedef pair<string,int>PSI;
typedef pair<string,string>PSS;
const int N=2e6+5,INF=0x3f3f3f3f,Mod=1e9+7,mod=998244353;
const double eps=1e-6;

void solve(){
    int n,a,b,aa,bb;cin>>n>>a>>b>>aa>>bb;
    int c=a*b/__gcd(a,b);
    int ac=n/a,bc=n/b,cc=n/c;
    if(aa>=bb)bc-=cc;
    else ac-=cc;
    cout<<ac*aa+bc*bb;
}
signed main(){
    ios::sync_with_stdio(0),cin.tie(0),cout.tie(0);
    int T=1;
    //cin>>T;
//    init();
    while(T--){
        solve();
    }
    return 0;
}
View Code

 

10.6

A - New Skateboard

思路：后两位能整除4的一定可以，枚举每个最后一位；还可以发现最后一位为0、4、8的只要前一位是偶数或最后一位是2、6的只要前一位是奇数也一定满足

#include<bits/stdc++.h>
using namespace std;
#define int long long
//#define int __int128
#define double long double
typedef pair<int,int>PII;
typedef pair<string,int>PSI;
typedef pair<string,string>PSS;
const int N=2e6+5,INF=0x3f3f3f3f,Mod=1e9+7,mod=998244353;
const double eps=1e-6;

void solve(){
    string s;cin>>s;
    int ans=0;
    for(int i=0;i<s.size();++i){
        if(s[i]=='0'||s[i]=='4'||s[i]=='8'){
            if(i&&(s[i-1]-'0')%2==0)ans+=i;
            ans++;
        }else if(s[i]=='2'||s[i]=='6'){
            if(i&&(s[i-1]-'0')%2)ans+=i;
        }
    }
    cout<<ans;
}
signed main(){
    ios::sync_with_stdio(0),cin.tie(0),cout.tie(0);
    int T=1;
    //cin>>T;
//    init();
    while(T--){
        solve();
    }
    return 0;
}
View Code

 

B - Two Seals

思路：枚举每对矩形，判断是否能装下即可

#include<bits/stdc++.h>
using namespace std;
#define int long long
//#define int __int128
#define double long double
typedef pair<int,int>PII;
typedef pair<string,int>PSI;
typedef pair<string,string>PSS;
const int N=2e6+5,INF=0x3f3f3f3f,Mod=1e9+7,mod=998244353;
const double eps=1e-6;

void solve(){
    int n,a,b,ans=0;cin>>n>>a>>b;
    vector<PII>ve(n);
    for(int i=0;i<n;++i){
        cin>>ve[i].first>>ve[i].second;
    }
    for(int i=0;i<n;++i){
        for(int j=0;j<i;++j){
            int x1=ve[i].first+ve[j].first,y1=max(ve[i].second,ve[j].second);
            int x2=ve[i].first+ve[j].second,y2=max(ve[i].second,ve[j].first);
            int x3=ve[i].second+ve[j].first,y3=max(ve[i].first,ve[j].second);
            int x4=ve[i].second+ve[j].second,y4=max(ve[i].first,ve[j].first);
            if(min(x1,y1)<=min(a,b)&&max(x1,y1)<=max(a,b))ans=max(ans,ve[i].first*ve[i].second+ve[j].first*ve[j].second);
            else if(min(x2,y2)<=min(a,b)&&max(x2,y2)<=max(a,b))ans=max(ans,ve[i].first*ve[i].second+ve[j].first*ve[j].second);
            else if(min(x3,y3)<=min(a,b)&&max(x3,y3)<=max(a,b))ans=max(ans,ve[i].first*ve[i].second+ve[j].first*ve[j].second);
            else if(min(x4,y4)<=min(a,b)&&max(x4,y4)<=max(a,b))ans=max(ans,ve[i].first*ve[i].second+ve[j].first*ve[j].second);
        }
    }
    cout<<ans;
}
signed main(){
    ios::sync_with_stdio(0),cin.tie(0),cout.tie(0);
    int T=1;
    //cin>>T;
//    init();
    while(T--){
        solve();
    }
    return 0;
}
View Code

 

C - The Same Calendar

思路：可以求出2016.1.1的星期，若n小于2016，那么先倒回去找到n的星期，再暴力往后枚举即可

#include<bits/stdc++.h>
using namespace std;
//#define int long long
//#define int __int128
#define double long double
typedef pair<int,int>PII;
typedef pair<string,int>PSI;
typedef pair<string,string>PSS;
const int N=2e6+5,INF=0x3f3f3f3f,Mod=1e9+7,mod=998244353;
const double eps=1e-6;

void solve() {
    int n;
    cin >> n;
    int p = 5, x , sx;
    if ((n % 4 == 0 && n % 100 != 0) || n % 400 == 0)sx = 366;
    else sx = 365;
    if (n == 2016)x = p;
    if (n < 2016) {
        for (int i = 2015;; --i) {
            int s = 365;
            if ((i % 4 == 0 && i % 100 != 0) || i % 400 == 0)s = 366;
            int c=(s-p+1)%7;
            if(c==0)c=7;
            p = 8 - c;
            if (i == n) {
                if(p==7)p=0;
                x = p;
                break;
            }
        }
    }
    int l=min(2016,n);
    for (int i = l;; ++i) {
        int s = 365;
        if ((i % 4 == 0 && i % 100 != 0) || i % 400 == 0)s = 366;
        p = (s + p) % 7;
        if (i + 1 == n)x = p;
        else if (i + 1 > n) {
            int ss=365;
            if (((i + 1) % 4 == 0 && (i + 1) % 100 != 0) || (i + 1) % 400 == 0)ss = 366;
            if (p == x && ss == sx) {
                cout << i + 1;
                break;
            }
        }
    }

}
signed main(){
    ios::sync_with_stdio(0),cin.tie(0),cout.tie(0);
    int T=1;
    //cin>>T;
//    init();
    while(T--){
        solve();
    }
    return 0;
}
View Code

 

10.7

 

A - New Skateboard GNU C++20 (64) Accepted 15 ms 1000 KB
227309026 Oct/09/2023 18:57UTC+8 bible_w B - Two Seal
posted @ 2023-09-27 20:17  bible_w  阅读(1)  评论(0)  编辑  收藏  举报