以下是引用beyondyf在2013-12-10 14:35:24的发言：

呵呵既然韶志给了这么高的评价，那就参与一下
#include <stdio.h>

void cal(int n, int a, int b)
{
    static int f[] = {0, 1, 2, 3, 4, 5, 6, 7, 8, 9};
    const int e[] = {1, 10, 100, 1000, 10000};
    int t, i;
    if(n == 5)
    {
        if(!(b % a) && (t = b / a) < 10)
            printf("%d * %d = %d\n", a, t, b);
        return;
    }
    for(i = n; i < 10; i++)
    {
        t = f[n]; f[n] = f; f = t;
        cal(n + 1, a * 10 + f[n], b + f[n] * e[n]);
        t = f[n]; f[n] = f; f = t;
    }
}

int main()
{
    cal(0, 0, 0);
    return 0;
}