读书-算法《程序设计导引及在线实践》-简单计算题1：鸡兔同笼

#include <stdio.h>
void main()
{
    int ncase, i, nfeet;
    scanf("%d", &ncase);
    for(i = 0; i < ncase; i++)
    {
        scanf("%d", &nfeet);
        if (nfeet%2 != 0)
        {
            printf("0 0\n");
        }
        else if (nfeet%4 != 0)
        {
            printf("%d %d",nfeet/4 + 1, nfeet/2);
        }
        else
        {
            printf("%d %d",nfeet/4, nfeet/2);
        }
    }
}