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Codeforces 821 A. Okabe and Future Gadget Laboratory

2017-06-26 14:45 417 查看
题意:

给定一个n*n的矩阵,判定A x,y=Ax,i+Aj,y (Ax,y <> 1)

算法:模拟

代码:

#include <bits/stdc++.h>

using namespace std;

typedef long long LL;
typedef double DB;
typedef unsigned int UI;
typedef pair<int, int> PII;

const int inf = 0x7f7f7f7f;

#define rdi() read<int>()
#define rdl() read<LL>()
#define rds(a) scanf("%s", a)
#define mk(i, j) make_pair(i, j)
#define pb push_back
#define fi first
#define se second
#define For(i, j, k) for (int i = j; i <= k; i ++)
#define Rep(i, j, k) for (int i = j; i >= k; i --)
#define Edge(i, u) for (int i = head[u]; i; i = e[i].nxt)

template<typename t> t read() {
t x = 0; int f = 1; char c = getchar();
while (c > '9' || c < '0') f = c == '-' ? -1 : 1 , c = getchar();
while (c >= '0' && c <= '9') x = x * 10 + c - 48 , c = getchar();
return x * f;
}

int n, a[100][100], fg;

int main() {
//  freopen(".in", "r", stdin);
//  freopen(".out", "w", stdout);
n = rdi();
for (int i = 1; i <= n; i ++) {
for (int j = 1; j <= n; j ++) a[i][j] = rdi();
}
for (int i = 1; i <= n; i ++) {
for (int j = 1; j <= n; j ++) if (a[i][j] != 1) {
fg = 0;
for (int x = 1; x <= n; x ++) {
for (int y = 1; y <= n; y ++) {
if (a[i][j] == a[i][x] + a[y][j]) fg = 1;
}
}
if (!fg) return puts("No"), 0;
}
}
puts("Yes");
return 0;
}
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