1205: C hicken and rabbit s (简单数学题)

本文探讨了鸡兔同笼问题的数学解决方案,通过给定的腿总数,计算笼中可能的鸡和兔数量范围。

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1205: C hicken and rabbit s

时间限制: 1 Sec   内存限制: 32 MB
提交: 53   解决: 36
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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
 
#include<stdio.h>
#include<string.h>
#include<math.h>
#include<algorithm>
#define MIN(x,y)(x<y?x:y)
#define MAX(x,y)(x>y?x:y)
using namespace std;
int main(){
	int t,n,min,max;
	scanf("%d",&t);
	while(t--)
	{
		min=0;max=0;
		scanf("%d",&n);
		if(n&1)
			printf("0 0\n");
		else
		{
			max=n/2;
			if(n%4==0)
				min=n/4;
			else
				min=(n-2)/4+1;
			printf("%d %d\n",min,max);
		}
	}
	return 0;
}

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