Codeforces 821 A. Okabe and Future Gadget Laboratory
2017-06-26 14:45
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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:
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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