CodeForces - 710C Magic Odd Square（奇数和幻方构造）

Magic Odd Square

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

题意：构造n阶幻方，填入1~n^2，使每行每列每对角线和为奇数。

思路：暴搜TLE，那么考虑构造。方法见下图，1填奇，0填偶。

     0 0 0 0 0 1 0 0 0 0 0
     0 0 0 0 1 1 1 0 0 0 0
     0 0 0 1 1 1 1 1 0 0 0
     0 0 1 1 1 1 1 1 1 0 0
     0 1 1 1 1 1 1 1 1 1 0
     1 1 1 1 1 1 1 1 1 1 1
     0 1 1 1 1 1 1 1 1 1 0
     0 0 1 1 1 1 1 1 1 0 0
     0 0 0 1 1 1 1 1 0 0 0
     0 0 0 0 1 1 1 0 0 0 0
     0 0 0 0 0 1 0 0 0 0 0

各类幻方构造法（知乎）：https://www.zhihu.com/question/30498489/answer/49208033

#include<bits/stdc++.h>
#define MAX 105
using namespace std;
typedef long long ll;

int a[MAX][MAX],b[MAX*MAX];

int main()
{
    int n,i,j,k;
    scanf("%d",&n);
    for(i=;i<=n/;i++){
        j=n/+-i+;
        for(k=;k<=*i-;k++){
            a[i][j+k-]=;
        }
    }
    for(i=n/+;i<=n;i++){
        j=i-(n/+)+;
        for(k=;k<=n+-(*j-);k++){
            a[i][j+k-]=;
        }
    }
    for(i=;i<=n;i++){
        for(j=;j<=n;j++){
            if(j>) printf(" ");
            if(a[i][j]){
                for(k=;k<=n*n;k+=){
                    if(b[k]) continue;
                    b[k]=;
                    a[i][j]=k;
                    break;
                }
            }
            else{
                for(k=;k<=n*n;k+=){
                    if(b[k]) continue;
                    b[k]=;
                    a[i][j]=k;
                    break;
                }
            }
            printf("%d",a[i][j]);
        }
        printf("\n");
    }
    return ;
 }