codeforces 710C Magic Odd Square(构造或者n阶幻方)

Find an n × n matrix with different numbers from 1 to n2, so the sum in each row, column and both main diagonals are odd.

Input

The only line contains odd integer n (1 ≤ n ≤ 49).

Output

Print n lines with n integers. All the integers should be different and from 1 to n2. The sum in each row, column and both main diagonals should be odd.

Examples

input

1

output

1

input

3

output

2 1 43 5 76 9 8分析：给你与1个奇数n, 让你构造一个n * n 的矩阵，要求保证这个矩阵的每行每列和主对角线上数字的和为奇数。构造：

 1 #include<bits/stdc++.h>
 2 using namespace std;
 3 int ans[50][50];
 4
 5 int main()
 6 {
 7     int n;
 8     ios::sync_with_stdio(false);
 9     cin.tie(0);
10     cin >> n;
11     memset(ans,0,sizeof(ans));
12     int num1 = 1,num2 =2,cnt1 = (n+1)/2,cnt2 = (n+1)/2;
13     for(int i = 1; i <= n; i++)
14     {
15         for(int j = 1; j <= n; j++)
16         {
17             if(abs(cnt1 - i) + abs(cnt2 - j) <= n/2) // 为什么这么写，有点不清楚
18                 ans[i][j] = num1,num1 += 2;
19             else
20                 ans[i][j] = num2,num2 += 2;
21         }
22     }
23     for(int i = 1; i <= n; i++)
24     {
25         for(int j = 1; j< n; j++)
26         {
27             printf("%d ",ans[i][j]);
28         }
29         printf("%d\n",ans[i][n]);
30     }
31     return 0;
32 }

n阶幻方：

 1 # include <stdio.h>
 2 # include <string.h>
 3 int g[50][50];
 4 int main(){
 5     int i, j, k, n, x, y, r, c;
 6     memset(g, 0, sizeof(g));
 7     scanf("%d", &n);
 8     r=1; c=(1+n)/2;
 9     g[1][c]=1;
10
11     for(i=2; i<=n*n; i++){
12         x=r;y=c;
13         x=x-1;
14         if(x<1){
15             x=n;
16         }
17         y=y+1;
18         if(y>n){
19             y=1;
20         }
21         if(g[x][y]){
22             g[r+1][c]=i;
23             r=r+1;
24         }
25         else{
26             g[x][y]=i;
27             r=x;c=y;
28         }
29     }
30     for(i=1; i<=n; i++){
31         for(j=1; j<=n; j++){
32             if(j!=n)
33             printf("%d ", g[i][j]);
34             else{
35                 printf("%d\n", g[i][j]);
36             }
37         }
38
39     }
40     return 0;
41 }