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[50][50];

int main()
{
    int n;
    ios::sync_with_stdio(false);
    cin.tie(0);
    cin >> n;
    memset(ans,0,sizeof(ans));
    int num1 = 1,num2 =2,cnt1 = (n+1)/2,cnt2 = (n+1)/2;
    for(int i = 1; i <= n; i++)
    {
        for(int j = 1; j <= n; j++)
        {
            if(abs(cnt1 - i) + abs(cnt2 - j) <= n/2) // 为什么这么写，有点不清楚
                ans[i][j] = num1,num1 += 2;
            else
                ans[i][j] = num2,num2 += 2;
        }
    }
    for(int i = 1; i <= n; i++)
    {
        for(int j = 1; j< n; j++)
        {
            printf("%d ",ans[i][j]);
        }
        printf("%d\n",ans[i][n]);
    }
    return 0;
}

n阶幻方：

# include <stdio.h>  
# include <string.h>  
int g[50][50];  
int main(){  
    int i, j, k, n, x, y, r, c;  
    memset(g, 0, sizeof(g));  
    scanf("%d", &n);  
    r=1; c=(1+n)/2;  
    g[1][c]=1;  
      
    for(i=2; i<=n*n; i++){  
        x=r;y=c;  
        x=x-1;  
        if(x<1){  
            x=n;  
        }  
        y=y+1;  
        if(y>n){  
            y=1;  
        }  
        if(g[x][y]){  
            g[r+1][c]=i;  
            r=r+1;  
        }  
        else{  
            g[x][y]=i;  
            r=x;c=y;  
        }  
    }  
    for(i=1; i<=n; i++){  
        for(j=1; j<=n; j++){  
            if(j!=n)  
            printf("%d ", g[i][j]);  
            else{  
                printf("%d\n", g[i][j]);  
            }  
        }  
          
    }  
    return 0;  
}