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 4
3 5 7
6 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：

 #include<bits/stdc++.h>
 using namespace std;
 int ans[][];

 int main()
 {
     int n;
     ios::sync_with_stdio(false);
     cin.tie();
     cin >> n;
     memset(ans,,sizeof(ans));
     int num1 = ,num2 =,cnt1 = (n+)/,cnt2 = (n+)/;
     for(int i = ; i <= n; i++)
     {
         for(int j = ; j <= n; j++)
         {
             if(abs(cnt1 - i) + abs(cnt2 - j) <= n/) // 为什么这么写，有点不清楚
                 ans[i][j] = num1,num1 += ;
             else
                 ans[i][j] = num2,num2 += ;
         }
     }
     for(int i = ; i <= n; i++)
     {
         for(int j = ; j< n; j++)
         {
             printf("%d ",ans[i][j]);
         }
         printf("%d\n",ans[i][n]);
     }
     return ;
 }

n阶幻方：

 # include <stdio.h>
 # include <string.h>
 int g[][];
 int main(){
     int i, j, k, n, x, y, r, c;
     memset(g, , sizeof(g));
     scanf("%d", &n);
     r=; c=(+n)/;
     g[][c]=;  

     for(i=; i<=n*n; i++){
         x=r;y=c;
         x=x-;
         if(x<){
             x=n;
         }
         y=y+;
         if(y>n){
             y=;
         }
         if(g[x][y]){
             g[r+][c]=i;
             r=r+;
         }
         else{
             g[x][y]=i;
             r=x;c=y;
         }
     }
     for(i=; i<=n; i++){
         for(j=; j<=n; j++){
             if(j!=n)
             printf("%d ", g[i][j]);
             else{
                 printf("%d\n", g[i][j]);
             }
         }  

     }
     return ;
 }