hdu4497 GCD and LCM

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

2013 ACM-ICPC吉林通化全国邀请赛——题目重现

 

 

Recommend

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 <iostream>

#include <stdio.h>

#include <algorithm>

#include <string.h>

#include <math.h>

using namespace std;

#define MAXN 100000

int 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<ans;i++)

            sum*=6*num[i];

        printf("%d\n",sum);

    }

    return 0;

}