Light OJ 1116 Ekka Dokka

最新推荐文章于 2016-08-21 21:31:20 发布

Grit_ICPC

最新推荐文章于 2016-08-21 21:31:20 发布

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分类专栏：数论文章标签：技巧 Lightoj

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本篇博客介绍了一个有趣的算法挑战：Ekka和Dokka希望分享一块蛋糕，根据他们对奇数和偶数的偏好，需要将蛋糕分为两部分，一部分面积为奇数N平方厘米，另一部分面积为偶数M平方厘米，且N*M等于预先设定的数值W。文章提供了输入输出样例，并给出了一段示例代码。

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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 andM. If there are multiple solutions, then print the result where M is as small as possible.

Sample Input

Output for Sample Input

Case 1: 5 2

Case 2: Impossible

Case 3: 3 4

#include<iostream>
#include<cstdio>
#include<cstring>
#include<cmath>
#define LL long long
using namespace std;
int main()
{
    LL n,m,i,x,y;
    LL cla;
    bool bj;
    scanf("%lld",&cla);
    for(int gr=1; gr<=cla; gr++)
    {
        bj=true;
        scanf("%lld",&n);
        printf("Case %lld: ",gr);
        i=1;
        while(n%2==0)//此题就是找出一个奇一个偶数
        {
            n=n/2;
            i*=2;
            bj=false;
        }
        if(!bj)//一个奇数，不可能分成一个奇数和一个偶数
            printf("%lld %lld\n",n,i);
        else
            printf("Impossible\n");
    }
    return 0;
}