Educational Codeforces Round 16 C. Magic Odd Square-优快云博客

本文探讨如何构造一个n×n的矩阵，该矩阵由从1到n²的不同整数组成，使得每一行、每一列及两条主对角线上的元素之和均为奇数。给出了一种简洁有效的解决方案。

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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其实这是一个很经典的问题啊，并不算简单，解决策略貌似也有很多种，这里贴几篇连接：点击打开链接 最后发现一个超级短的解决办法，但是原理不是很懂，改一改就过了#include <stdio.h>
#include <stdlib.h>
#include <cstdio>
int f(int n, int x, int y)
{
    return (x + y*2 + 1)%n;
}

int main(int argc, char **argv)
{
    int i, j,n;
    scanf("%d",&n);
    for (i = 0; i < n; i++) {
        for (j = 0; j < n; j++)
        {
           printf("%d ", f(n, n - j - 1, i)*n + f(n, j, i) + 1);
        }
        putchar('\n');
    }
    
    return 0;
}