LIGHTOJ-1116 - Ekka Dokka

最新推荐文章于 2019-02-13 23:29:27 发布

原创最新推荐文章于 2019-02-13 23:29:27 发布 · 1k 阅读

0 ·

CC 4.0 BY-SA版权

文章标签：

#LIGHTOJ #1116 #Ekka Dokka

数学同时被 2 个专栏收录

20 篇文章

订阅专栏

LIGHTOJ--ACM

7 篇文章

订阅专栏

本文探讨了如何通过数学方法解决一个趣味性的蛋糕分配问题，即如何将一个蛋糕按照一个人喜欢的奇数平方厘米和另一个人喜欢的偶数平方厘米进行公平分配，满足特定的乘积条件。

1116 - Ekka Dokka

PDF (English)

Statistics

Forum

Time Limit: 2 second(s)

Memory Limit: 32 MB

Ekka and his friend Dokka decided to buy a cake. They both love cakes and that's why they want to share the cake after buying it. As the name suggested that Ekka is very fond of odd numbers and Dokka is very fond of even numbers, they want to divide the cake such that Ekka gets a share of N square centimeters and Dokka gets a share of M square centimeters where N is odd and M is even. Both N and M are positive integers.

They want to divide the cake such that N * M = W, where W is the dashing factor set by them. Now you know their dashing factor, you have to find whether they can buy the desired cake or not.

Input

Input starts with an integer T (≤ 10000), denoting the number of test cases.

Each case contains an integer W (2 ≤ W < 2⁶³). And W will not be a power of 2.

Output

For each case, print the case number first. After that print "Impossible" if they can't buy their desired cake. If they can buy such a cake, you have to print N and M. If there are multiple solutions, then print the result whereM is as small as possible.

Sample Input

Output for Sample Input

Case 1: 5 2

Case 2: Impossible

Case 3: 3 4

让一个数分成一奇数，一偶数的乘积！

#include<cstdio>
#include<cstring>
#include<cmath>
#include<algorithm>
using namespace std;
int main()
{
	long long a,b,i,j,m,x,y,sum,cot=1;
	scanf("%lld",&m);
	while(m--)
	{
		scanf("%lld",&a);
		if(a&1)
		{
			printf("Case %lld: Impossible\n",cot++);
			continue;
		}
		b=a;
		sum=1;
		while(b%2==0)
		{
			b/=2;
			if(a%(long long)pow(2,sum)==0)
			{
				x=a/(long long)pow(2,sum);
				y=(long long)pow(2,sum);
			}
			sum++;
		}
		printf("Case %lld: %lld %lld\n",cot++,x,y);
	}
	return 0;
}