GCD and LCM

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65535/65535 K (Java/Others)

Total Submission(s): 78 Accepted Submission(s): 43

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

 

 

Source

 

 

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liuyiding

很明显，m/n!=0的话，就直接输出0就可以了！否刚，直接分解质因数m/n,找到，每个质因子的个数，这样，我们，就可以得出每个质因数为a1^k1,那是题目就是要把这k1个a1分到三个数中，那么排列组合就是k1*A（3，2）,也就是，6*k1,种，直接算出来就行了！

#include 
     
      #include 
      
       #include 
       
        #include 
        
         #include 
         
          using namespace std;#define MAXN 100000int num[MAXN];int main(){    int tcase,n,m,i,ans,sum,tempm;    scanf("%d",&tcase);    while(tcase--)    {        scanf("%d%d",&n,&m);        memset(num,0,sizeof(num));        if(m%n!=0)        {            printf("0\n");            continue;        }        m=m/n;tempm=sqrt(m)+1;        for(i=2,ans=0;i<=tempm;i++)        {            if(m%i==0)            {                while(m%i==0)                {                    num[ans]++;m/=i;                }                ans++;            }        }        if(m!=1)            num[ans++]=1;        for(sum=1,i=0;i