[C++知识库](a/GCD) * (b/GCD) = LCM/GCD

GCD & LCM
Given x and y (2 <= x <= 100,000, 2 <= y <= 1,000,000), you are to count the number of p and q such that:

1. p and q are positive integers;
2. GCD(p, q) = x;
3. LCM(p, q) = y.

Input

x and y, one line for each test.

Output

Number of pairs of p and q.

Sample Input

3 60

Sample Output

4

a * b=GCD * LCM
=> (a/GCD) * (b/GCD) = LCM/GCD
寻找乘积为 LCM/GCD 且互质的两个数
一来从 LCM 到 LCM/GCD缩小了查找范围，二来寻找一对互质数可以作为筛查条件

其实一个思想很重要：把求解的数转化成和质数有关来求解，因为质数相对合数少很多，所以这样处理问题范围减小很多！
验证素数的时候只需验证到该数的开方根，降低复杂度

#include <iostream>
using namespace std;
int gcd(int a,int b){
	if(a%b==0)return b;
	return gcd(b,a%b);
}
int main(){
//	GCD(p, q) = x,LCM(p, q) = y --> p*q==x*y
	int x,y;
	while(cin>>x>>y){
		int res=0;
		if(y%x!=0){
			cout<<"0"<<endl;
			continue;
		}
		else{
			int m=y/x;
			for(int i=1;i<=m;i++){//寻找乘积为m且互质的两个数 
				if(m%i==0){//i m/i
					if(gcd(i,m/i)==1)res++;
				} 
			}
		}
		cout<<res<<endl;
	} 
    return 0;
}