GCD and LCM HDU

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.
Input
First 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.
Output
For 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
这个记得黑书上有唯一分解定理形式的对gcd和lcm的解释，然后就想起来了，我能做出来的题，一般都是简单题…

#include<cstdio>#include<iostream>#include<algorithm>#include<cstring>#include<cmath>#include<set>#include<vector>#define N 200005#define mod 1000000007using namespace std;bool p[N];vector<int> prime;void init(){    for(int i=2;i<N;i++)        if(!p[i])        {            prime.push_back(i);            for(int j=i+i;j<N;j+=i)                p[j]=true;        }}int fac[500];int e[500];int cnt;void getFac(long long x){    cnt=0;    for(int i=0;(long long)prime[i]*prime[i]<=x;i++)    {        if(x%prime[i]==0)        {            fac[cnt]=prime[i];            e[cnt]=0;            while(x%prime[i]==0)            {                x/=prime[i];                e[cnt]++;            }            cnt++;        }    }            if(x>1)        {            fac[cnt]=x;            e[cnt]=1;            cnt++;        }}int main(){    int t;    init();    scanf("%d",&t);    long long G,L;    while(t--)    {        scanf("%lld%lld",&G,&L);        if(L%G)            printf("0\n");        else        {            L/=G;            getFac(L);            long long ans=1;            for(int i=0;i<cnt;i++)                ans*=6*e[i];            printf("%lld\n",ans);        }    }    return 0;}