数学趣题——爱因斯坦阶梯问题

若x mod 2 =1, x mod 3 = 2, x mod 5 = 4, x mod 6 = 5, x mod 7 =0;求最小解

 

源码如下：

#include <string.h>
#include <stdio.h>

int main()
{
    int x = 7, i, res, flag = 0;

    for (i=1; i<=100; i++)
    {
        if ((x%2 == 1) && (x % 3 == 2) && (x % 5 == 4) && (x % 6 ==5))
        {
            res = x;
            flag = 1;
            break;
        }
        x = 7 * (i + 1);
    }

    if (flag == 1)
        printf("the result is %d", res);
    else
        printf("no result");

    return 0;
}