Question

用c语言编程：鸡兔同笼，有三十五只头，九十四个脚，求鸡兔各有多少？

用穷举法求解，九十鸡或兔的个数x从0到35循环试探，如果满足条件的x就是一个解，有多个解。

Answer 1

#include <stdio.h>
void main()
{
int chicken, rabbit;
printf("鸡\t兔\n");
for (chicken = 0; chicken <= 35; chicken ++)
{
rabbit = 35 - chicken;
if (8 * rabbit + 2 * chicken == 94)
printf("%d\t%d\n", chicken, rabbit);
}
}

追问

这是枚举法？

追答

是啊，不过略微改良了一下，其实只有一组解，完全意义上的枚举法如下：
#include
int main()
{
int chicken, rabbit;
printf("鸡\t兔\n");
for (chicken = 0; chicken <= 35; chicken ++)
{
for (rabbit =0; rabbit <= 35; rabbit ++)
{
if (rabbit + chicken == 35 && 4 * rabbit + 2 * chicken == 94)
printf("%d\t%d\n", chicken, rabbit);
}
}
}

Answer 2

#include <stdio.h>
void main()
{
int chicken, rabbit;
for (chicken = 0; chicken <= 35; chicken ++)
{
rabbit = 35 - chicken;
if (4 * rabbit + 2 * chicken == 94)
printf("鸡:%d 兔%d\n", chicken, rabbit);
}
}