2023/5/14 Study Summary

In this question, we can see that the data range is very small, so we can use violent enumeration to traverse all the lengths and widths that can form a rectangle. At the same time, we must meet the condition that there are no obstacles in this rectangle to obtain the maximum perimeter.

#include<bits/stdc++.h>
using namespace std;
typedef long long ll;
string mp[30];
bool solve(int i ,int j ,int k,int l){
	for(int m=i;m<=k;m++){
		for(int n=j;n<=l;n++){
			if(mp[m][n]=='1') return false;
		}
	}
	return true;
}
int main()
{
	int n,m;
	cin>>n>>m;
	for(int i=0;i<n;i++){
			cin>>mp[i];
	}
	int res=0;
	for(int i=0;i<n;i++){
            for(int j=0;j<m;j++){
                for(int k=i;k<n;k++){
                    for(int l=j;l<m;l++){
                        if(solve(i,j,k,l)&&((k-i+1)*2+2*(l-j+1))>res){
                            res=(k-i+1)*2+2*(l-j+1);
                        }
                    }
                }
            }
     }
     cout<<res<<endl;
}

First of all, the special judgment first, when m=1 and n=1 and n<=m, it may cause the game to never end, and then we traverse

2~m, if n%i can be satisfied, the game may never end.

#include<bits/stdc++.h>
using namespace std;
typedef long long ll;
#define maxn 1000000

int main()
{
	int t;
	cin>>t;
	while(t--){
		int n;
		cin>>n;
		int f=0;
		if(n==1||m==1){
			cout<<"YES"<<endl;
		}else if(m>=n) cout<<"NO"<<endl;
		 else{
		for(int i=2;i<=m&&i<=n/i;i++){
			if(n%i==0){
				f=1;
			}
		}
		if(f==0) cout<<"YES"<<endl;
		else cout<<"NO"<<endl;
	}
}
}