Problem Description

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

程式碼及註釋如下：
 

/*
此題的G和L範圍在int範圍內,所以求其中的最大素因子不超過100000;
所以第一步需要求出1-10萬中的素數,然後儲存到vector陣列中...
第二步就是求個數了..
如果LCM不能除盡GCD的話,說明不能組成,直接輸出0.
如果能夠除盡,L需要先除以G,將兩個數的公共素因子冪全部消去...
最後是素因子分解L.
每分解一個素因子就用6*素因子的指數....
答案就是素因子的指數乘6累成起來的...
*/
#include <cstdio>
#include <cstring>
#include <algorithm>
#include <iostream>
#include <vector>
using namespace std;
const int maxn=100000;
vector<int>pri;
int isnot[maxn+5];
int t;
int x,y;
int Size;
void Judge ()
{
    memset (isnot,0,sizeof(isnot));
    isnot[1]=1;
    for (int i=2;i<=maxn;i++)
    {
        if(!isnot[i])
        {
            pri.push_back(i);
            for (int j=2*i;j<=maxn;j+=i)
                isnot[j]=1;
        }
    }
    Size=pri.size();
}
int main()
{
    Judge();
    scanf("%d",&t);
    while (t--)
    {
        int num=1;
        scanf("%d%d",&x,&y);
        if(y%x)
            printf("0\n");
        else
        {
            y/=x;
            for (int i=0;i<Size;i++)
            {
                int cnt=0;
                if(pri[i]*pri[i]>y)
                    break;
                while(y%pri[i]==0)
                {
                    cnt++;
                    y/=pri[i];
                }
                if(cnt)
                    num*=6*cnt;
            }
            if(y!=1)
                num*=6;
            printf("%d\n",num);
        }
    }
    return 0;
}