POJ 2429 GCD & LCM Inverse（素数判定Miller-Rabin+素因子分解Pollard-rho+二进制枚举）

GCD & LCM Inverse

Time Limit: 2000MS		Memory Limit: 65536K
Total Submissions: 18825		Accepted: 3486

Description

Given two positive integers a and b, we can easily calculate the greatest common divisor (GCD) and the least common multiple (LCM) of a and b. But what about the inverse? That is: given GCD and LCM, finding a and b.

Input

The input contains multiple test cases, each of which contains two positive integers, the GCD and the LCM. You can assume that these two numbers are both less than 2^63.

Output

For each test case, output a and b in ascending order. If there are multiple solutions, output the pair with smallest a + b.

Sample Input

3 60

Sample Output

12 15

Source

POJ Achilles

算法分析：

题意：

给出gcd(a,b)和lcm(a,b)，求a和b，如果存在多组方案则输出a+b最小的那一组

分析：

首先，我们可以看出(x/gcd)*(y/gcd)=lcm/gcd，并且x/gcd和y/gcd互质，那么我们先利用把所有的质数求出来Pollard_Rho，将相同的质数合并，现在的问题转变成把合并后的质数分为两堆，使得x+y最小，然后我们可以采用二进制枚举（解释点这里），一个个枚举出来。

代码实现：

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