HDU 4497 GCD and LCM

E - GCD and LCM

Time Limit:1000MS     Memory Limit:65535KB     64bit IO Format:%I64d & %I64u

Submit  Status  Practice  HDU 4497

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 
        
        
        
        

 

//公式推导
#include <cstdio>
#include <cstring>
#include <algorithm>
using namespace std;
const int maxn=1000000+5;
int primes[maxn];
int T,G,L;
int main() {
    //get_prime();
    scanf("%d",&T);
    while(T--) {
        scanf("%d%d",&G,&L);
        if(L%G)
        {
            printf("0\n");
            continue;
        }
        int sk=L/G;
        //deal_prime(sk);
        memset(primes,0,sizeof(primes));
        int tot=0;
        for(int i=2; i<=sk; i++) {
            if(sk%i==0) {
                while(sk%i==0) {
                    primes[tot]++;
                    sk/=i;
                }
                //printf("[%d]\n",primes[tot]);
                tot++;
            }
        }
        if(sk>1)primes[tot++]++;
        int res=1;
        for(int i=0; i<tot; i++) {
            if(primes[i])
                res*=(primes[i]*6);
        }
        printf("%d\n",res);
    }
    return 0;
}