Codeforces Round #420 (Div. 2) A. Okabe and Future Gadget Laboratory
2017-06-26 12:52
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题目大意
给出一个n*n矩阵,判断对于每个不等于1的数是否存在其所在行和所在列数相加等于这个数。example
31 1 2
2 3 1
6 4 1
output:Yes
2=1+1,
2=1+1,3=2+1,
6=2+4.4=3+1.
题解
暴力枚举即可。#include<iostream> #include<cstdio> #include<cstring> #include<algorithm> using namespace std; int read() { char ch=getchar();int f=0; while(ch<'0'||ch>'9') ch=getchar(); while(ch>='0'&&ch<='9') {f=(f<<1)+(f<<3)+ch-'0'; ch=getchar();} return f; } int a[55][55]; int main() { int n=read(); for(int i=1;i<=n;i++) { for(int j=1;j<=n;j++) a[i][j]=read(); } for(int i=1;i<=n;i++) { for(int j=1;j<=n;j++) { if(a[i][j]!=1) { bool ac=0; for(int k=1;k<=n;k++) { if(k==j) continue; for(int l=1;l<=n;l++) { if(i==l) continue; if(a[i][k]+a[l][j]==a[i][j]) { ac=1; break; } } if(ac==1) { break; } } if(!ac) { printf("No"); return 0; } } } } printf("Yes"); return 0; }
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