CodeForces 821A Okabe and Future Gadget Laboratory
2017-07-11 12:45
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题目链接:http://codeforces.com/contest/821/problem/A
题意:给你一个n*n的矩阵,问你对于矩阵里每一个不等于1的元素a[x][y]来说,是否存在s和t使得a[x][s]+a[t][y] = a[x][y],如果对于每一个都存在,那么输出yes,否则输出no
解析:由于n比较小只有50,所以直接暴力枚举就好
题意:给你一个n*n的矩阵,问你对于矩阵里每一个不等于1的元素a[x][y]来说,是否存在s和t使得a[x][s]+a[t][y] = a[x][y],如果对于每一个都存在,那么输出yes,否则输出no
解析:由于n比较小只有50,所以直接暴力枚举就好
#include <bits/stdc++.h> using namespace std; const int inf = 0x7fffffff; const int maxn = 100; int a[maxn][maxn]; int main(void) { int n; scanf("%d",&n); for(int i=1;i<=n;i++) { for(int j=1;j<=n;j++) scanf("%d",&a[i][j]); } int flag = 1; for(int i=1;i<=n;i++) { for(int j=1;j<=n;j++) { if(a[i][j]!=1) { int tt = 0; for(int k=1;k<=n;k++) { for(int l=1;l<=n;l++) { if(k!=j && l!=i) { if(a[i][k]+a[l][j]==a[i][j]) { tt=1; break; } } } if(tt) break; } if(!tt) { flag = 0; break; } } } if(!flag) break; } if(flag) puts("Yes"); else puts("No"); return 0; }
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