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 (SoungEarlf@gmail.com)
* @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
", ans?"Yes":"No");
}
return 0;
}