#include <stdio.h>
#include <stdlib.h>
int main()
{
int leg,n,i;
int Max,Min;
scanf("%d",&n);
for(i=0; i<n; i++)
{
scanf("%d",&leg);
if(leg%2!=0)//当该定数之不成立时,及鸡兔的腿不可能是奇数
{
Min=0;
Max=0;
}
else
{
if(leg%4==0)//腿的数目是4的倍数及可能完全是鸡或兔
{
Max=leg/2;
Min=leg/4;
}
else//否则这个该定的偶数只需减去一个鸡的腿及为兔的腿数目
{
Max=leg/2;
Min=(leg-2)/4+1;
}
}
printf("%d %d\n",Min,Max);
}
return 0;
}
题目描述
Chicken and rabbits are in a same cage. As we all know, chicken has two legs but rabbit has four. Now
we know the number of legs in the cage is A, please tell me how many animals may in the cage at least
and at most.
输入
The first line of the input contains the number of test cases in the file. Each test case that follows
consists of one lines. each case contains only one integer numbers A specifying the total legs in the
cage .
输出
For each test case, print a line contains the answer
样例输入
2
3
20
样例输出
0 0 5 10
本文通过C语言实现了一种解决经典的鸡兔同笼问题的方法。程序接收一个表示笼中动物腿总数的整数,并输出笼中最少和最多的动物数量。通过简单的逻辑判断和数学计算,能够有效地处理各种情况,包括当腿数为奇数或4的倍数时的特殊情况。
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