Codeforces Round #420 A
阿新 • • 發佈:2017-07-01
clu set tac 一個 nbsp 思路 turn 不為 stack
Okabe and Future Gadget Laboratory
題意:給一個矩陣,如果矩陣中任意一個不為1的數都滿足 存在一對 s t 使得 ais+atj=aij
思路:xjb暴力寫
AC代碼:
#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<<x<<" "<<"UUUUU"<<endl; #define mem(a) memset(a,0,sizeof(a)) using namespace std; const int N=1e5+100; int a[55][55],n; int check(int l,int r){ for(int i=1; i<=n; ++i){ for(int j=1; j<=n; ++j){ if(a[l][i]+a[j][r]==a[l][r])return 1; } } return 0; } int main(){ cin>>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; } /* 3 1 1 2 2 3 1 6 4 1 3 1 5 2 1 1 1 1 2 3 */
Codeforces Round #420 A