题意：给一个矩阵，如果矩阵中任意一个不为1的数都满足 存在一对 s t 使得 ais+atj=aij

思路：xjb暴力写

AC代码：

#include "iostream"#include "string.h"#include "stack"#include "queue"#include "string"#include "vector"#include "set"#include "map"#include "algorithm"#include "stdio.h"#include "math.h"#define ll long long#define bug(x) cout<
     
      <<" "<<"UUUUU"<
      
       >n;    for(int i=1; i<=n; ++i){        for(int j=1; j<=n; ++j){            cin>>a[i][j];        }    }    for(int i=1; i<=n; ++i){        for(int j=1; j<=n; ++j){            if(a[i][j]!=1){                if(!check(i,j)){                    cout<<"No";                    return 0;                }            }        }    }    cout<<"Yes";    return 0;}/*31 1 22 3 16 4 131 5 21 1 11 2 3*/