Ekka Dokka（水题）

最新推荐文章于 2020-11-20 17:37:09 发布

原创最新推荐文章于 2020-11-20 17:37:09 发布 · 696 阅读

0 ·

CC 4.0 BY-SA版权

文章标签：

#numbers #64bit #input #each #io

本篇博客介绍了一个有趣的算法题目——EkkaDokka蛋糕分配问题。该问题要求将一块蛋糕按照特定条件（N为奇数，M为偶数且N*M等于给定值W）分成两部分。文章提供了详细的输入输出样例和实现代码。

Ekka Dokka

Time Limit:2000MS Memory Limit:32768KB 64bit IO Format:%lld & %llu

Submit Status

Description

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 Mare 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

Sample Output

Case 1: 5 2

Case 2: Impossible

Case 3: 3 4

注：n&1==1表明n为奇数

#include <iostream>
#include <cstdio>
using namespace std;

int main()
{
    int T, ca = 1;
    long long w, m, n, i, j;
    scanf("%d", &T);
    while(T--)
    {
        cin >> w;
        if(w&1) printf("Case %d: Impossible\n", ca++);
        else
        {
            for(m = 2; m < w; m += 2)
            {
                n = w / m;
                if(n * m == w && n&1)
                    break;
            }
            if(m < w) cout << "Case " << ca++ << ": " << n << " " << m << endl;
            else printf("Case %d: Impossible\n", ca++);
        }
    }
    return 0;
}