hdu 4497 GCD and LCM 数学

GCD and LCM

Time Limit: 20 Sec

Memory Limit: 256 MB

题目连接

http://acm.hdu.edu.cn/showproblem.php?pid=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

HINT

题意

问你有多少个三元组，可以使得他们lcm等于b,gcd等于a

题解：

和二元组一样做，显然直接b/a之后，分解质因数，然后直接跑排列组合就好了

代码：

//qscqesze
#include <cstdio>
#include <cmath>
#include <cstring>
#include <ctime>
#include <iostream>
#include <algorithm>
#include <set>
#include <vector>
#include <sstream>
#include <queue>
#include <typeinfo>
#include <fstream>
#include <map>
#include <stack>
typedef long long ll;
using namespace std;
//freopen("D.in","r",stdin);
//freopen("D.out","w",stdout);
#define sspeed ios_base::sync_with_stdio(0);cin.tie(0)
#define test freopen("test.txt","r",stdin)
#define maxn 200001
#define mod 10007
#define eps 1e-9
int Num;
char CH[];
const int inf=0x3f3f3f3f;
const ll infll = 0x3f3f3f3f3f3f3f3fLL;
inline ll read()
{
    ll x=,f=;char ch=getchar();
    while(ch<''||ch>''){if(ch=='-')f=-;ch=getchar();}
    while(ch>=''&&ch<=''){x=x*+ch-'';ch=getchar();}
    return x*f;
}
inline void P(int x)
{
    Num=;if(!x){putchar('');puts("");return;}
    while(x>)CH[++Num]=x%,x/=;
    while(Num)putchar(CH[Num--]+);
    puts("");
}
//**************************************************************************************

ll ans[maxn];
int main()
{
    //test;
    int t=read();
    while(t--)
    {
        memset(ans,,sizeof(ans));
        ll a,b;
        scanf("%lld%lld",&a,&b);
        if(b%a)
        {
            cout<<""<<endl;
            continue;
        }
        ll c=b/a;
        ll kiss=;
        int tot=;
        for(int i=;i<=c;i++)
        {
            if(c==)
                break;
            if(c%i==)
            {
                while(c%i==)
                    ans[tot]++,c/=i;
                tot++;
            }
        }
        for(int i=;i<tot;i++)
        {
            if(ans[i])
                kiss*=ans[i]*;
        }
        cout<<kiss<<endl;
    }
}