Ekka Dokka LightOJ 1116简单数学

最新推荐文章于 2019-02-13 16:37:23 发布

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本文介绍了一个有趣的数学问题：如何将一块蛋糕按照特定条件分成两份，一份为奇数平方厘米，另一份为偶数平方厘米，并确保两份的乘积等于预先设定的数值。文章详细解释了解决这一问题的基本思路和算法实现。

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

题意：将 $\;W\;$ 写成 $\;W=N*M\;$ 的形式，其中 $N$ 是奇数， $M$ 是偶数， $M$ 要最小的，如果不能够写成这种形式，输出 $Impossible$
由算数基本定理可以得到

W = 2 a * p a 1 1 * p a 2 2 * p a 3 3 \dots * p a n n

$W=2^a*p_1^{a_1}*p_2^{a_2}*p_3^{a_3}\cdots*p_n^{a_n}$
因为质数除了2之外，其他的全部是奇数
所以

M = 2 a

$M=2^a$

N = p a 1 1 * p a 2 2 * p a 3 3 \dots * p a n n = W 2 a

$N=p_1^{a_1}*p_2^{a_2}*p_3^{a_3}\cdots*p_n^{a_n}=\frac{W}{2^a}$
当

a=0a=0 $a=0$ 的时候

MM $M$ 不存在，输出

I m p o s s i b l e

$Impossible$
基本思路就是这样

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;
}