Ekka Dokka LightOJ 1116简单数学

Dear 丶 2022-05-18 00:07 121阅读 0赞

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**  
3

**Sample Output**  
Case 1: 5 2

Case 2: Impossible

Case 3: 3 4

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题意：将W  W 写成W=N∗M  W = N ∗ M 的形式，其中N  N 是奇数，M M是偶数，M  M 要最小的，如果不能够写成这种形式，输出Impossible I m p o s s i b l e  
由算数基本定理可以得到

W=2a∗pa11∗pa22∗pa33⋯∗pann  W = 2 a ∗ p 1 a 1 ∗ p 2 a 2 ∗ p 3 a 3 ⋯ ∗ p n a n

因为质数除了2之外，其他的全部是奇数  
所以

M=2a  M = 2 a

N=pa11∗pa22∗pa33⋯∗pann=W2a  N = p 1 a 1 ∗ p 2 a 2 ∗ p 3 a 3 ⋯ ∗ p n a n = W 2 a

当  a=0  a = 0  的时候  M  M  不存在，输出  Impossible I m p o s s i b l e  
基本思路就是这样

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code：

#include<bits/stdc++.h>
    using namespace std;
    typedef long long ll;
    int main()
    {
        int t;
        cin>>t;
        for(int k=1;k<=t;k++)
        {
            ll n=1,m=1;
            ll w;
            scanf("%lld",&w);
            if(w%2==0)
            {
                while(w%2==0)
                {
                    n*=2;
                    w/=2;
                }
                printf("Case %d: %lld %lld\n",k,w,n);
            }
            else    
                printf("Case %d: Impossible\n",k);
        }
        return 0;
    }