light oj 1116 - Ekka Dokka

1116 - Ekka Dokka

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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 where M is as small as possible.

Sample Input	Output for Sample Input
3 10 5 12	Case 1: 5 2 Case 2: Impossible Case 3: 3 4

暴力可以直接过  需要注意的是 n和m都为整数且n为奇数m为偶数

#include<stdio.h>
#include<string.h>
#include<stdlib.h>
#include<algorithm>
#include<math.h>
#define LL long long
#define DD double
#define MAX 10000
using namespace std;
int main()
{
	int t;
	int m;
	LL w,n,i,j;
	int ans;
	DD ant;
	scanf("%d",&t);
	int k=1;
	while(t--)
	{
		scanf("%lld",&w);
		int flag=0;
		printf("Case %d: ",k++);
		if(w&1)
		{
			printf("Impossible\n");
			continue;
		}
		else
		{
			for(i=2;i<=w/3+4;i++)
			{
				if(i&1)
				continue;
				if(w%i==0)
				{
					n=w/i;
					if(n&1)
					{
						flag=1;
						break;
					}
				}
			}
			if(flag)
			printf("%lld %lld\n",n,i);
			else
			printf("Impossible\n");
		}
	}
	return 0;
}