王朝 鸡兔同笼求鸡兔总和最大最小问题

#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