CodeForces 711B Chris and Magic Square (暴力，水题)

题意：给定n*n个矩阵，其中只有一个格子是0，让你填上一个数，使得所有的行列的对角线的和都相等。

析：首先n为1，就随便填，然后就是除了0这一行或者这一列，那么一定有其他的行列是完整的，所以，先把其他的算出来，然后再作差就算这个数了，

然后再去验证其他的对不对就好了。除了n为1，其他的都是唯一解应该。或者没有。

代码如下：

#pragma comment(linker, "/STACK:1024000000,1024000000")
#include <cstdio>
#include <string>
#include <cstdlib>
#include <cmath>
#include <iostream>
#include <cstring>
#include <set>
#include <queue>
#include <algorithm>
#include <vector>
#include <map>
#include <cctype>
#include <cmath>
#include <stack>
#include <list>
#include <sstream>
#define frer freopen("in.txt", "r", stdin)
#define frew freopen("out.txt", "w", stdout)
using namespace std;

typedef long long LL;
typedef pair<int, int> P;
const int INF = 0x3f3f3f3f;
const double inf = 0x3f3f3f3f3f3f;
const double PI = acos(-1.0);
const double eps = 1e-8;
const int maxn = 5e2 + 5;
const int mod = 1e9 + 7;
const int dr[] = {-1, 1, 0, 0, 1, 1, -1, -1};
const int dc[] = {0, 0, 1, -1, 1, -1, 1, -1};
const char *de[] = {"0000", "0001", "0010", "0011", "0100", "0101", "0110", "0111", "1000", "1001", "1010", "1011", "1100", "1101", "1110", "1111"};
int n, m;
const int mon[] = {0, 31, 28, 31, 30, 31, 30, 31, 31, 30, 31, 30, 31};
const int monn[] = {0, 31, 29, 31, 30, 31, 30, 31, 31, 30, 31, 30, 31};
inline int Min(int a, int b){ return a < b ? a : b; }
inline int Max(int a, int b){ return a > b ? a : b; }
inline LL Min(LL a, LL b){ return a < b ? a : b; }
inline LL Max(LL a, LL b){ return a > b ? a : b; }
inline bool is_in(int r, int c){
    return r >= 0 && r < n && c >= 0 && c < m;
}
LL a[maxn][maxn];

int main(){
    while(cin >> n){
        memset(a, 0, sizeof a);
        LL sum1 = 0, sum2 = 0;
        int x, y;
        for(int i = 0; i < n; ++i){
            for(int j = 0; j < n; ++j){
                scanf("%I64d", &a[i][j]);
                a[i][n] += a[i][j];
                a[n][j] += a[i][j];
                if(a[i][j] == 0)  x = i, y = j;
                if(i == j)  sum1 += a[i][j];
                if(i + j == n - 1)  sum2 += a[i][j];
            }
        }
        if(1 == n)  printf("1\n");
        else{
            LL ans = 0;
            for(int i = 0; i < n; ++i){
                if(i != x){
                    ans = a[i][n] - a[x][n];
                    a[x][n] += ans;
                    a[n][y] += ans;
                    if(x == y)  sum1 += ans;
                    if(x + y == n - 1)  sum2 += ans;
                    break;
                }
            }
            bool ok = true;
            if(sum1 != sum2 || ans < 1 || ans > 1e18) ok = false;
            if(!ok){  printf("-1\n"); continue;  }
            for(int i = 0; i < n; ++i){
                if(a[i][n] != sum1 || a[n][i] != sum1){  ok = false; break; }
            }
            if(!ok){  printf("-1\n"); continue;  }
            else cout << ans << endl;
        }
    }
    return 0;
}