Codeforces Round 871 (Div. 4)

发布时间 2023-05-07 12:49:19作者: bible_w

Codeforces Round 871 (Div. 4)

A - Love Story

#include<bits/stdc++.h>
using namespace std;
typedef pair<int,int>PII;
typedef pair<string,int>PSI;
const int N=1e5+5,INF=0x3f3f3f3f,Mod=1e6;

const double eps=1e-6;
typedef long long ll;

int n;
int main(){
    ios::sync_with_stdio(0),cin.tie(0),cout.tie(0);
    string s;
    cin>>n;
    string a="codeforces";
    while(n--){
        cin>>s;
        int c=0;
        for(int i=0;i<10;++i){
            if(s[i]!=a[i])c++;
        }
        cout<<c<<'\n';
    }
    return 0;
}
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B - Blank Space

#include<bits/stdc++.h>
using namespace std;
typedef pair<int,int>PII;
typedef pair<string,int>PSI;
const int N=1e5+5,INF=0x3f3f3f3f,Mod=1e6;

const double eps=1e-6;
typedef long long ll;

int t,n;
int main(){
    ios::sync_with_stdio(0),cin.tie(0),cout.tie(0);
    cin>>t;
    while(t--){
        cin>>n;
        int res=0;
        for(int i=0,x,ma=0;i<n;++i){
            cin>>x;
            if(x==0)ma++;
            else ma=0;
            res=max(res,ma);
        }
        cout<<res<<'\n';
    }

    return 0;
}
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C - Mr. Perfectly Fine

#include<bits/stdc++.h>
using namespace std;
typedef pair<int,int>PII;
typedef pair<string,int>PSI;
const int N=1e5+5,INF=0x3f3f3f3f,Mod=1e6;

const double eps=1e-6;
typedef long long ll;

int t,n;
int main(){
    ios::sync_with_stdio(0),cin.tie(0),cout.tie(0);
    cin>>t;
    while(t--){
        cin>>n;
        int l=INF,r=INF,lp=-1,rp=-1,a=INF;
        string s;
        for(int i=0,m;i<n;++i){
            cin>>m>>s;
            if(s[0]=='1'&&s[1]=='1')a=min(m,a);
            else{
                if(s[0]=='1')l=min(l,m),lp=i;
                if(s[1]=='1')r=min(r,m),rp=i;
            }
        }
        if((lp==-1||rp==-1)&&a==INF)cout<<"-1\n";
        else{
            cout<<min(a,l+r)<<'\n';
        }
    }

    return 0;
}
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D - Gold Rush

#include<bits/stdc++.h>
using namespace std;
typedef pair<int,int>PII;
typedef pair<string,int>PSI;
const int N=1e7+5,INF=0x3f3f3f3f,Mod=1e6;

const double eps=1e-6;
typedef long long ll;

int t,n,m;
bool st[N];
bool bfs(){
    memset(st,false,sizeof st);
    queue<int>q;
    q.push(n);
    st[n]=true;
    while(q.size()){
        int f=q.front();q.pop();
        if(f==m){
            return true;
        }
        if(f%3==0){
            if(!st[f/3*2])q.push(f/3*2),st[f/3*2]=true;
            if(!st[f/3])q.push(f/3),st[f/3]=true;
        }
    }
    return false;
}
int main(){
    ios::sync_with_stdio(0),cin.tie(0),cout.tie(0);
    cin>>t;
    while(t--){
        cin>>n>>m;
        if(n>=m&&bfs())cout<<"YES\n";
        else cout<<"NO\n";
    }

    return 0;
}
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E - The Lakes

#include<bits/stdc++.h>
using namespace std;
typedef pair<int,int>PII;
typedef pair<string,int>PSI;
const int N=1e3+5,INF=0x3f3f3f3f,Mod=1e6;

const double eps=1e-6;
typedef long long ll;

int t,n,m;
int a[N][N];
int dx[4]={-1,0,1,0},dy[4]={0,1,0,-1};
int bfs(int x,int y){
    int res=a[x][y];
    queue<PII>q;
    q.push({x,y});
    a[x][y]=0;
    while(q.size()){
        auto f=q.front();q.pop();
        int x1=f.first,y1=f.second;
        for(int i=0;i<4;++i){
            int xx=x1+dx[i],yy=y1+dy[i];
            if(xx>=0&&xx<n&&yy>=0&&yy<m&&a[xx][yy]>0){
                res+=a[xx][yy];
                q.push({xx,yy});
                a[xx][yy]=0;
            }
        }
    }
    return res;
}
int main(){
    ios::sync_with_stdio(0),cin.tie(0),cout.tie(0);
    cin>>t;
    while(t--){
        cin>>n>>m;
        int res=0;
        for(int i=0;i<n;++i)
            for(int j=0;j<m;++j)cin>>a[i][j];
        for(int i=0;i<n;++i){
            for(int j=0;j<m;++j){
                if(a[i][j]>0){
                    res=max(res,bfs(i,j));
                }
            }
        }
        cout<<res<<'\n';
    }

    return 0;
}
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F - Forever Winter

思路:求每个节点的入度,除了叶子节点的入度为1,剩余的入度只有两种或一种;若为两种,数量少的为中心

#include<bits/stdc++.h>
using namespace std;
typedef pair<int,int>PII;
typedef pair<string,int>PSI;
const int N=2e2+5,INF=0x3f3f3f3f,Mod=1e6;

const double eps=1e-6;
typedef long long ll;

int t,n,m;
int ru[N];
int main(){
    ios::sync_with_stdio(0),cin.tie(0),cout.tie(0);
    cin>>t;
    while(t--){
        cin>>n>>m;
        memset(ru,0,sizeof ru);
        for(int i=0,u,v;i<m;++i){
            cin>>u>>v;
            ru[u]++,ru[v]++;
        }
        map<int,int>mp;
        for(int i=1;i<=n;++i){
            if(ru[i]>1){
                mp[ru[i]]++;
            }
        }
        if(mp.size()==1){
            int c=mp.begin()->first;
            cout<<c<<' '<<c-1<<'\n';
        }
        else{
            int a[2],b[2],i=0;
            for(auto u:mp){
                a[i]=u.first;
                b[i++]=u.second;
            }
            if(b[0]<b[1])swap(b[0],b[1]),swap(a[0],a[1]);
            cout<<a[1]<<' '<<a[0]-1<<'\n';
        }
    }

    return 0;
}
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G - Hits Different

思路:a[i][j]表示第i行第j列的数,a[i][j]=a[i-1][j]+a[i-1][j-1]-a[i-2][j-1],(有被重复加的格子);第i层的数有i*(i+1)/2个,前i层的数有i*(i+1)*(i+2)/6个,可以二分求出层数,也可以直接用数组标记每个数的位置

#include<bits/stdc++.h>
using namespace std;
typedef pair<int,int>PII;
typedef pair<string,int>PSI;
const int N=1500+5,INF=0x3f3f3f3f,Mod=1e6;

const double eps=1e-6;
typedef long long ll;

ll t,n,a[N],idx;
ll b[N][N];
void P(){
    ll i;
    for(i=1;;++i){
        a[i]=i*(i+1)/2;
        if(a[i]>=1e6)break;
    }idx=i;
    ll c=1;
    for(ll k=1;k<=idx;++k){
        for(ll j=1;j<=k;++j){
            b[k][j]=c*c;c++;
            b[k][j]+=b[k-1][j]+b[k-1][j-1];
            if(k>2&&j!=1&&j!=k)b[k][j]-=b[k-2][j-1];
        }
    }
    
}

int main(){
    ios::sync_with_stdio(0),cin.tie(0),cout.tie(0);
    cin>>t;
    P();
    while(t--){
        cin>>n;
        ll p= lower_bound(a+1,a+idx+1,n)-a;
        ll i=p,j=n-a[p-1];
        cout<<b[i][j]<<'\n';
    }
    return 0;
}
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