CF821 A. Okabe and Future Gadget Laboratory 水

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题意：询问n X n中非1数是否能够由同行同列中分别取两个数做和得到。

思路：水题。

/** @Date    : 2017-07-03 16:23:18
  * @FileName: A.cpp
  * @Platform: Windows
  * @Author  : Lweleth ([email protected])
  * @Link    : https://github.com/
  * @Version : $Id$
  */
#include <bits/stdc++.h>
#define LL long long
#define PII pair
#define MP(x, y) make_pair((x),(y))
#define fi first
#define se second
#define PB(x) push_back((x))
#define MMG(x) memset((x), -1,sizeof(x))
#define MMF(x) memset((x),0,sizeof(x))
#define MMI(x) memset((x), INF, sizeof(x))
using namespace std;

const int INF = 0x3f3f3f3f;
const int N = 1e5+20;
const double eps = 1e-8;

int n;
int a[60][60];
int main()
{
	while(cin >> n)
	{
		for(int i = 1; i <= n; i++)
		{
			for(int j = 1; j <= n; j++)
				scanf("%d", &a[i][j]);
		}
		int ans = 1;
		for(int i = 1; i <= n && ans; i++)
		{
			for(int j = 1; j <= n && ans; j++)
			{
				if(a[i][j] == 1)
					continue;
				int flag = 1;
				for(int k = 1; k <= n; k++)
				{
					if(j == k || a[i][k] >= a[i][j])
						continue;
					if(!flag)
						break;
					for(int l = 1; l <= n; l++)
					{
						if(!flag)
							break;
						if(l == i || a[i][k] + a[l][j] != a[i][j])
							continue;
						else
						{
							flag = 0;
							break;
						}
					}
				}
				if(flag)
				{
					//cout << i << j << endl;
					ans = 0;
					break;
				}

			}
		}
		printf("%s\n", ans?"Yes":"No");
	}
    return 0;
}