题解，较简便

#include <iostream>

using namespace std;

const int N = 40;

int n;
int a[N][N];

int main()
{
    cin >> n;

    int x = 0, y = n / 2;
    for (int i = 1; i <= n * n; i ++ )
    {
        a[x][y] = i;
        int tx = (x - 1 + n) % n, ty = (y + 1) % n;
        if (a[tx][ty])
            x ++ ;
        else
        {
            x = tx, y = ty;
        }
    }

    for (int i = 0; i < n; i ++ )
    {
        for (int j = 0; j < n; j ++ )
        {
            printf("%d%c", a[i][j], " \n"[j == n - 1]);
        }
    }

    return 0;
}