B. Chris and Magic Square
time limit per test2 seconds
memory limit per test256 megabytes
inputstandard input
outputstandard output
ZS the Coder and Chris the Baboon arrived at the entrance of Udayland. There is a n × n magic grid on the entrance which is filled with integers. Chris noticed that exactly one of the cells in the grid is empty, and to enter Udayland, they need to fill a positive integer into the empty cell.

Chris tried filling in random numbers but it didn’t work. ZS the Coder realizes that they need to fill in a positive integer such that the numbers in the grid form a magic square. This means that he has to fill in a positive integer so that the sum of the numbers in each row of the grid (), each column of the grid (), and the two long diagonals of the grid (the main diagonal — and the secondary diagonal — ) are equal.

Chris doesn’t know what number to fill in. Can you help Chris find the correct positive integer to fill in or determine that it is impossible?

Input
The first line of the input contains a single integer n (1 ≤ n ≤ 500) — the number of rows and columns of the magic grid.

n lines follow, each of them contains n integers. The j-th number in the i-th of them denotes ai, j (1 ≤ ai, j ≤ 109 or ai, j = 0), the number in the i-th row and j-th column of the magic grid. If the corresponding cell is empty, ai, j will be equal to 0. Otherwise, ai, j is positive.

It is guaranteed that there is exactly one pair of integers i, j (1 ≤ i, j ≤ n) such that ai, j = 0.

Output
Output a single integer, the positive integer x (1 ≤ x ≤ 1018) that should be filled in the empty cell so that the whole grid becomes a magic square. If such positive integer x does not exist, output  - 1 instead.

If there are multiple solutions, you may print any of them.

Examples
input
3
4 0 2
3 5 7
8 1 6
output
9
input
4
1 1 1 1
1 1 0 1
1 1 1 1
1 1 1 1
output
1
input
4
1 1 1 1
1 1 0 1
1 1 2 1
1 1 1 1
output
-1
Note
In the first sample case, we can fill in 9 into the empty cell to make the resulting grid a magic square. Indeed,

The sum of numbers in each row is:

4 + 9 + 2 = 3 + 5 + 7 = 8 + 1 + 6 = 15.

The sum of numbers in each column is:

4 + 3 + 8 = 9 + 5 + 1 = 2 + 7 + 6 = 15.

The sum of numbers in the two diagonals is:

4 + 5 + 6 = 2 + 5 + 8 = 15.

In the third sample case, it is impossible to fill a number in the empty square such that the resulting grid is a magic square.

題意：
給一個n*n的矩陣，裡面有且只有一個0，其他都是正整數，先求是否存在一個可以填在0位置的數，使得矩陣的每行之和與每列之和與兩個對角線之和都相等。

題解：
直接求值後check就可以了，然而FST了，因為資料中有除了0所在的行、列外，其他本身就不符合的行/列。

//
//  main.cpp
//  cf-369-2-2
//
//  Created by 袁子涵 on 16/8/29.
//  Copyright © 2016年 袁子涵. All rights reserved.
//

#include <iostream>
#include <cstdio>
#include <cstring>
#include <cstdlib>
#include <cmath>
#include <algorithm>
#include <queue>
#include <set>
#include <stack>
#include <string>
#include <map>

using namespace std;
const int maxn=505;
long long int n,num[maxn][maxn],x,y,nowx,total,tmp,tmp1,tmp2;
int main(int argc, const char * argv[]) {
    cin >> n;
    for (int i=1; i<=n; i++) {
        for (int j=1; j<=n; j++) {
            cin >> num[i][j];
            if (num[i][j]==0) {
                x=i,y=j;
            }
        }
    }
    if (n==1) {
        cout << 1 << endl;
        return 0;
    }
    if (x==1) {
        nowx=n;
    }
    else
    {
        nowx=1;
    }
    total=0;
    for (int i=1; i<=n; i++) {
        total+=num[nowx][i];
    }
    tmp=0;
    for (int i=1; i<=n; i++) {
        if (i==y) {
            continue;
        }
        else
        {
            tmp+=num[x][i];
        }
    }
    if (total-tmp<=0) {
        cout << -1 << endl;
        return 0;
    }
    num[x][y]=total-tmp;
    tmp=0;
    bool flag=1;
    for (int i=1; i<=n; i++) {
        tmp+=num[i][y];
    }
    if (tmp!=total) {
        flag=0;
    }
    tmp=0;
    for (int i=1; i<=n; i++) {
        tmp+=num[i][i];
    }
    if (tmp!=total) {
        flag=0;
    }
    tmp=0;
    for (int i=1; i<=n; i++) {
        tmp+=num[i][n-i+1];
    }
    if (tmp!=total) {
        flag=0;
    }
    for (int i=1; i<=n; i++) {
        tmp1=tmp2=0;
        for (int j=1; j<=n; j++) {
            tmp1+=num[i][j];
            tmp2+=num[j][i];
        }
        if (tmp1!=tmp2 || tmp1!=total || tmp2!=total) {
            flag=0;
            break;
        }
    }
    if (flag) {
        cout << num[x][y] << endl;
    }
    else
        cout << -1 << endl;
    return 0;
}