魔方阵

群里的朋友问了这道题，刚开始想到的最笨了：枚举，然后就是判断，要是n过大了就。。。。

后来参考百度百科给出的分类考虑的方法，尝试了一下，感觉按照方法来一点也不复杂了。。

不过百度百科也提到了一种通用的方法，没有尝试。

#include<stdio.h>

# define MAXN 100

int a[MAXN][MAXN];

void swap(int *x, int *y)
{
    int tmp;
    tmp = *x;
    *x = *y;
    *y = tmp;
}

void print_odd(int n)
{
    int i, j, t;
    i = 1;
    j = (n+1)/2;
    for (t = 1; t <= n*n; ++t)
    {
        a[i][j] = t;
        if (t%n == 0) ++i;
        else
        {
            if ((--i) == 0) i = n;
            if ((++j) == (n+1)) j = 1;
        }
    }
}

void print_even4(int n)
{
    int i, j, p, q, t, tmp, c;
    t = 0;
    for (i = 1; i <= n; ++i)
        for (j = 1; j <= n; ++j)
            a[j][i] = ++t;
    c = n/4;
    for (p = 1; p <= c; ++p)
        for (q = 1; q <= c; ++q)
        {
            i = p*4 - 3;
            j = q*4 - 3;
            swap(&a[i][j], &a[i+3][j+3]);
            swap(&a[i+1][j+1], &a[i+2][j+2]);
            swap(&a[i+3][j], &a[i][j+3]);
            swap(&a[i+2][j+1], &a[i+1][j+2]);
        }
}

void print_even2(int n)
{
    int i, j, c, u, m;
    c = n/2;
    u = c*c;
    print_odd(c);
    for (i = 1; i <= c; ++i)
        for (j = 1; j <= c; ++j)
        {
            a[i+c][j+c] = a[i][j]+u;
            a[i+c][j] = a[i][j]+3*u;
            a[i][j+c] = a[i][j]+2*u;
        }
    m = (c-1)/2;
    for (i = 1; i <= c; ++i)
        for (j = n-m+1; j <= n; ++j)
        {
            swap(&a[i][j], &a[i+c][j]);
            if (i != m+1) swap(&a[i][j-n+m], &a[i+c][j-n+m]);
            else swap(&a[i][j-n+m+m-1], &a[i+c][j-n+m+m-1]);
        }
}

int main()
{
    int i, j, t, n;
    while (~scanf("%d", &n))
    {
        if (n%2 == 1) print_odd(n);
        else if (n%4 == 0) print_even4(n);
        else print_even2(n);
        for (i = 1; i <= n; ++i)
        {
            for (j = 1; j <=n; ++j)
                printf("%4d ", a[i][j]);
            printf("\n");
        }
    }
    return 0;
}