A

模拟

#include<bits/stdc++.h>

using namespace std;

#define int long long
#define all(x) x.begin()+1,x.end()

void solve(){
    int n;
    cin>>n;
    std::vector<int> a(n+1),b(n+1);
    for(int i=1;i<=n;i++){
        cin>>a[i];
    }
    for(int i=1;i<=n;i++){
        cin>>b[i];
        if(b[i]<a[i]){
            cout<<"YES"<<endl;
            return ;
        }
    }
    cout<<"NO"<<endl;
    return ;

}

signed main(){
    int t=1;
    while(t--) solve();
    return 0;

}

B

经典的DFS求连通块的问题

#include<bits/stdc++.h>

using namespace std;

#define int long long
#define all(x) x.begin()+1,x.end()

const int N=110;

char a[N][N];
int n,m;
int dx[4]={1,-1,0,0};
int dy[4]={0,0,-1,1};

void dfs(int x,int y){
    for(int i=0;i<4;i++){
        int x1=dx[i]+x;
        int y1=dy[i]+y;
        if(x1<1||x1>n||y1<1||y1>m){
            continue;
        }
        if(a[x1][y1]!='B'){
            continue;
        }
        else{

            a[x1][y1]='.';
            dfs(x1,y1);
        }
    }
}

void solve(){
    cin>>n>>m;
    for(int i=1;i<=n;i++){
        for(int j=1;j<=m;j++){
            cin>>a[i][j];
        }
    }
    int ans=0;

    for(int i=1;i<=n;i++){
        for(int j=1;j<=m;j++){
            if(a[i][j]=='B'){
                ans++;
                dfs(i,j);

            }
        }
    }
    cout<<ans<<endl;
    return ;

}

signed main(){
    int t=1;
    while(t--) solve();
    return 0;

}

C

一般求求最大值最小值往往是可以用二分解决的

#include<bits/stdc++.h>

using namespace std;

#define int long long
#define all(x) x.begin()+1,x.end()

const int N=110;

int n,m,x,y;

bool check(int h){
    int ans1=h/x;
    int ans2=h/y;
    int ans3=h/(x*y);
    int maxn1=max(0ll,n-(ans2-ans3));
    int maxn2=max(0ll,m-(ans1-ans3));
    return maxn1+maxn2<=h-ans1-ans2+ans3;

}

void solve(){
    cin>>n>>m>>x>>y;
    int l=0;
    int r=1e9+10;
    while(l<r){
        int mid=(l+r)/2;
        if(check(mid)){
            r=mid;
        }
        else{
            l=mid+1;

        }
    }
    cout<<l<<endl;
    return ;

}

signed main(){
    int t=1;
    while(t--) solve();
    return 0;

}