lightoj 1116 - Ekka Dokka 【分解因子】

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 Nand 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

 

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PROBLEM SETTER: MUHAMMAD RIFAYAT SAMEE

SPECIAL THANKS: JANE ALAM JAN

题意：给定一个数W，问你能否找到一个奇数N和一个偶数M使得N*M = W，若存在则输出M最小时的一组解，反之输出Impossible。

思路：W必须是偶数。可以每次将W的二进制向右移一位，M的二进制向左移动一位，直到W为奇数为止。

AC代码：

#include <cstdio>
#include <cstring>
#include <cmath>
#include <cstdlib>
#include <algorithm>
#include <queue>
#include <stack>
#include <map>
#include <vector>
#define INF 0x3f3f3f3f
#define eps 1e-8
#define MAXN 500000+10
#define MAXM 50000000
#define Ri(a) scanf("%d", &a)
#define Rl(a) scanf("%lld", &a)
#define Rs(a) scanf("%s", a)
#define Pi(a) printf("%d\n", (a))
#define Pl(a) printf("%lld\n", (a))
#define Ps(a) printf("%s\n", (a))
#define W(a) while(a--)
#define CLR(a, b) memset(a, (b), sizeof(a))
#define MOD 1000000007
#define LL long long
using namespace std;
int main()
{
    int t, kcase = 1;
    Ri(t);
    W(t)
    {
        LL W; Rl(W);
        if(W & 1)
        {
            printf("Case %d: Impossible\n", kcase++);
            continue;
        }
        bool flag = false;
        LL N = 1, M = 1;
        for(;;)
        {
            M <<= 1;
            W >>= 1;
            if(W & 1)
            {
                N = W;
                flag = true;
                break;
            }
            if(W == 0)
                break;
        }
        if(!flag)
            printf("Case %d: Impossible\n", kcase++);
        else
            printf("Case %d: %lld %lld\n", kcase++, N, M);
    }
    return 0;
}