cf369 B Chris and Magic Square

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.
题意：使每行每列，左斜，右斜相加的和都相等，找出那个0的数为几。
我wa了3次终于找全了问题。
第一：n为1的情况，直接输出1；
例如：
1
0
第二：
3
3 8 1
2 4 6
7 0 5
输出的ans不能等于0
第三：
3
61 0 41
11 31 51
21 71 1
输出的ans应该大于0

#include<stdio.h>#include<math.h>#include<string.h>#include<stdlib.h>#include<iostream>#include<algorithm>using namespace std;#include<map>#include<queue>#include<vector>const int maxn=505;long long int ll[maxn];//行相加long long int rr[maxn];//竖着相加int a[maxn][maxn];long long int l,r;//斜着long long int li,rj;//标记0的位置long long int ans;int main(){    int n;    while(~scanf("%d",&n))    {        memset(ll,0,sizeof(ll));        memset(rr,0,sizeof(rr));        memset(a,0,sizeof(a));        l=0,r=0;        li=0,rj=0;        ans=0;        int flag=1,flag2=0,flag3=0,flag4=0;        //flag表示是否进行下一步，flag2标记左斜是否存在0        //flag3标记右斜是否存在0，flag4标记是否找到0的位置        for(int i=0; i<n; i++)        {            for(int j=0; j<n; j++)            {                scanf("%d",&a[i][j]);                if(a[i][j]==0)                {                    li=i,rj=j;                    flag4=1;                }                ll[i]+=a[i][j];                rr[j]+=a[i][j];                if(i==j)                {                    l+=a[i][j];                    if(flag4&&li==i&&rj==j)                    {                        flag2=1;                    }                }                if(j==n-i-1)                {                    r+=a[i][j];                    if(flag4&&li==i&&rj==n-i-1)                    {                        flag3=1;                    }                }            }        }        if(n==1)//特判1        {            printf("1\n");        }        else        {            for(int i=0; i<n; i++)//求ans            {                if(i!=li)                {                    ans=ll[i]-ll[li];                    ll[li]+=ans;                    break;                }            }            rr[rj]+=ans;            if(flag2)                l+=ans;            if(flag3)                r+=ans;            long long int fans=ll[0];            for(int i=1; i<n; i++)            {                if(fans!=ll[i])                {                    flag=0;                    break;                }            }            for(int i=0; i<n; i++)            {                if(fans!=rr[i])                {                    flag=0;                    break;                }            }            if(flag)            {                if(l!=fans)                {                    flag=0;                }            }            if(flag)            {                if(r!=fans)                {                    flag=0;                }            }            if(flag&&ans>0)            {                printf("%lld\n",ans);            }            else            {                printf("-1\n");            }        }    }    return 0;}