Given two positive integers G and L, could you tell me how many solutions of (x, y, z) there are, satisfying that gcd(x, y, z) = G and lcm(x, y, z) = L?
Note, gcd(x, y, z) means the greatest common divisor of x, y and z, while lcm(x, y, z) means the least common multiple of x, y and z.
Note 2, (1, 2, 3) and (1, 3, 2) are two different solutions.
Note, gcd(x, y, z) means the greatest common divisor of x, y and z, while lcm(x, y, z) means the least common multiple of x, y and z.
Note 2, (1, 2, 3) and (1, 3, 2) are two different solutions.
InputFirst line comes an integer T (T <= 12), telling the number of test cases.
The next T lines, each contains two positive 32-bit signed integers, G and L.
It’s guaranteed that each answer will fit in a 32-bit signed integer.OutputFor each test case, print one line with the number of solutions satisfying the conditions above.Sample Input
2 6 72 7 33
Sample Output
72 0昨天想了好久都没想通，，今天早上灵感突然的就来了。题解：首先我们要理解，最大公约数和最小公倍数的关系，比如说a*b=gcd（a,b）*lcm(a,b) 如果两边同时除以gcd的平方 lcm%gcd==0 所以，如果lcm%gcd!=0的话 应该不会存在关系第二： 我们让gcd和lcm同时除以gcd可得新的gcd和lcm gcd=1 lcm=lcm/gcd 他们分别是x/gcd y/gcd z/gcd的gcd和lcm 因此我们只要分解lcm/gcd就可以了第三：lcm=p1^max(a1,a2,a3)*p2^max(b1,b2,b3).... x0=p1^a1.... y0=p1^b1... z0=p1^c1...假如说a1 b1 c1 都不为0，那么他们的最大公约数不会是1，因此他们三者中至少有一个为0 ，最多有两个为0（3个为0的情况不会出现，p1一定是其中一个数的只质因子）由于是有顺序的(0,a1,c1)加入最多的为a1那么C1的取值为0--a1我们先考虑为0和相等的情况 有a1-1中，，变换一下顺序一共有6（a1-1）种,还有(0,0,a1)和（0,a1,a1）我们没考虑共3+3种因此共有6（a1-1）+6种
#include<iostream>
#include<cstdio>
using namespace std;
const int N=1e6+7;
bool p[N]={1,1,0};
int prime[N];
int k=0;
void pre(){
k=0;
for(int i=2;i<N;i++){
if(p[i]==0){
prime[k++]=i;
for(int j=i+i;j<=N;j+=i){
p[i]=1;
}
}
}
}
int main(){
pre();
int t;
cin>>t;
while(t--){
int n,m;
scanf("%d%d",&n,&m);
if(m%n!=0){//最大公约数应该是最下公倍数的系数比如说a*b=gcd*lcm两边同时除以gcd的平方，so lcm%gcd=0
puts("0");
continue ;
}
int x=m/n;
int sum=1;
for(int i=0;i<k&&prime[i]<x;i++){
if(x%prime[i]==0){
int ans=0;
while(x%prime[i]==0){
ans++;
x=x/prime[i];
}
sum*=6*ans;
}
}
if(x>1) sum*=6;
cout<<sum<<endl;
}
return 0;
}