hdu 4497 GCD and LCM

GCD and LCM

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65535/65535 K (Java/Others)
Total Submission(s): 1092    Accepted Submission(s): 512

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

 

题意:求满足 gcd(x,y,z) = g , lcm(x，y,z) = l  的 组数

首先 l%g != 0 肯定是不行的，

我们对 l 和 g 经行素数分解

g = p1^a1 * p2 ^ a2 * p3 ^ a3 ...

l = p1^b1 * p2 ^ b2 * p3 ^ b3 ...

因为 l %g == 0 ;所以 ai <= bi 

对于p1 素数，

对于 x,y,z ，需要有一个数的素数分解 中 p1的次数为 ai ；

一个数的素数分解 中 p1的次数为 bi 

如果 ai == bi ，那么 没有选择

ai < bi 的选择就是 C(1,3)*C(1,2)*(bi-a1+1) 公式就是 6*(bi-ai+1) 

意思就是，三个里面选一个取 ai，然后两个里面选一个选 bi ，第三个的任意选

#include<iostream>
#include<cstdio>
#include<cstring>
#include<algorithm>
#include<cmath>
#include<queue>
#include<vector>
#include<set>
#include<stack>
#include<map>
#include<ctime>
#include<bitset>
#define LL long long
#define mod 1000000007
#define maxn 110
#define pi acos(-1.0)
#define eps 1e-8
#define INF 0x3f3f3f3f
using namespace std;

bool check(LL i )
{
    for(int j = 2 ; j*j <= i ;j++)if(i%j == 0)
        return false;
    return true;
}
int main()
{
    int j,i,l,g;
    int T,ans1,u,v;
    LL ans;
    cin >> T ;
    while(T--)
    {
        scanf("%d%d",&g,&l ) ;
        if(l%g !=0)puts("0") ;
        else
        {
            l /= g ;
            ans=1;
            for( LL i = 2 ; i*i <= l ;i++)if(l%i==0&&check(i))
            {
                v = 0;
                while(l%i==0)
                {
                    v++;
                    l /= i ;
                }
                ans *= v*6 ;
            }
            if(l !=1) ans *= 6 ;
            printf("%I64d
",ans);
        }
    }
    return 0 ;
}