鸡兔同笼-题解（C语言代码，思路清晰，简单易懂）

解题思路:设鸡和兔子的数量为x,y,则有x+y=n,2x+4y=m,即可得x=(4n-m)/2,y=(m-2n)/2，只有x,y为分数，或者为负数时，即为无解情况。详细代码如下：

#include<stdio.h>

int main()

{

    double n,m,chicken_num,rabbit_num;

    while(scanf("%lf%lf",&n,&m) == 2)

    {

        chicken_num = (4*n-m)/2;

        rabbit_num = (m-2*n)/2;

        if(chicken_num != (int)chicken_num || rabbit_num != (int)rabbit_num || chicken_num < 0 || rabbit_num < 0)

            printf("No answer\n");

        else

            printf("%.0lf %.0lf\n",chicken_num,rabbit_num);

    }

    return 0;

}