Codeforces Gym100735 E.Restore (KTU Programming Camp (Day 1) Lithuania, Birˇstonas, August 19, 2015)

 

E - Restore

 

Given a matrix A of size N * N. The rows are numbered from 0 to N-1, the columns are numbered from 0 to N-1. In this matrix, the sums of each row, the sums of each column, and the sum of the two diagonals are equal.

For example, a matrix with N = 3:

The sums of each row:

2 + 9 + 4 = 15

7 + 5 + 3 = 15

6 + 1 + 8 = 15

The sums of each column:

2 + 7 + 6 = 15

9 + 5 + 1 = 15

4 + 3 + 8 = 15

The sums of each diagonal:

2 + 5 + 8 = 15

4 + 5 + 6 = 15

As you can notice, all sums are equal to 15.

However, all the numbers in the main diagonal (the main diagonal consists of cells (i, i)) have been removed. Your task is to recover these cells.

Input

The first line contains the dimension of the matrix, n (1 ≤ n ≤ 100).

The following n lines contain n integers each, with removed numbers denoted by 0, and all other numbers  - 10¹² ≤ A_ij ≤ 10¹².

Output

The restored matrix should be outputted in a similar format, with n lines consisting of n integers each.

Example

Input

3
0 9 4
7 0 3
6 1 0

Output

2 9 4 
7 5 3 
6 1 8 

 

这个题有一个地方有点坑，就是如果一行里面有两个0，因为只有对角线的挖掉了，所以直接竖着算这一排的就可以。其他的没了。

 

代码:

 1 #include<iostream>
 2 #include<cstring>
 3 #include<cstdio>
 4 #include<algorithm>
 5 using namespace std;
 6 typedef long long ll;
 7 ll a[105][105];
 8 int main(){
 9     int n;
10     while(~scanf("%d",&n)){
11             ll cnt=0;
12         for(int i=0;i<n;i++){
13             for(int j=0;j<n;j++){
14                 scanf("%lld",&a[i][j]);
15                 cnt+=a[i][j];
16             }
17         }
18         cnt/=n-1;int i,j;   //算那个相等的值
19         for(i=0;i<n;i++){
20                 ll num=0;int flag,ret=0;
21             for(j=0;j<n;j++){
22                 num+=a[i][j];
23                 if(a[i][j]==0){flag=j;ret++;}
24             }
25             if(ret>1){    //如果这一排有多于1个0
26                 ll qwe=0;
27                 for(int k=0;k<n;k++)qwe+=a[k][i];
28                 a[i][i]=cnt-qwe;
29             }
30             else if(num!=cnt)a[i][flag]=cnt-num;
31         }
32         for(int i=0;i<n;i++){
33             for(int j=0;j<n;j++){
34                 printf("%lld",a[i][j]);
35                 if(j!=n-1)printf(" ");
36                 else printf("\n");
37             }
38         }
39     }
40     return 0;
41 }