HDU - 4497 GCD and LCM 数论gcd

传送门

题意：

给三个数的$lcm$和$gcd$，求满足条件的三元组组合个数。

思路：

首先$lcm\mathrm{m}\mathrm{o}\mathrm{d}gcd==0$是有组合的条件，否则输出0。
现在可知$lcm(x^{{}^{\prime }},y^{{}^{\prime }},z^{{}^{\prime }})=\frac{lcm(x,y,z)}{gcd(x,y,z)}，gcd(x^{{}^{\prime }},y^{{}^{\prime }},z^{{}^{\prime }})=1$，对$a$分解质因子得到$p_1^{u_1}{p}_2^{u_2}...p_n^{u_n}$，假设$x^{{}^{\prime }}=p_1^{i_1}{p}_2^{i_2}...p_n^{j_n},y^{{}^{\prime }}=p_1^{j_1}{p}_2^{j_2}...p_n^{j_n},z^{{}^{\prime }}=p_1^{k_1}{p}_2^{k_2}...p_n^{k_n}$。那么由于$gcd(x^{{}^{\prime }},y^{{}^{\prime }},z^{{}^{\prime }})=1$，可知$min(i_1,j_1,k_1)=0$,$max(i_1,j_1,k_1)=u_1$，所以我们需要找出来一个位置取$0$，一个位置取$u_1$，其他的位置随意就好了。当前位置的答案即为$A_3^2\ast u_1=6\ast u_1$，那么$ans=\sum 6\ast u_i$。

    //#pragma GCC optimize(2)
#include<cstdio>
#include<iostream>
#include<string>
#include<cstring>
#include<map>
#include<cmath>
#include<cctype>
#include<vector>
#include<set>
#include<queue>
#include<algorithm>
#include<sstream>
#include<ctime>
#include<cstdlib>
#define X first
#define Y second
#define L (u<<1)
#define R (u<<1|1)
#define pb push_back
#define mk make_pair
#define Mid (tr[u].l+tr[u].r>>1)
#define Len(u) (tr[u].r-tr[u].l+1)
#define random(a,b) ((a)+rand()%((b)-(a)+1))
#define db puts("---")
using namespace std;

//void rd_cre() { freopen("d://dp//data.txt","w",stdout); srand(time(NULL)); }
//void rd_ac() { freopen("d://dp//data.txt","r",stdin); freopen("d://dp//AC.txt","w",stdout); }
//void rd_wa() { freopen("d://dp//data.txt","r",stdin); freopen("d://dp//WA.txt","w",stdout); }

typedef long long LL;
typedef unsigned long long ULL;
typedef pair<int,int> PII;

const int N=1000010,mod=1e9+7,INF=0x3f3f3f3f;
const double eps=1e-6;

LL g,l;

int main()
{
    
    
//	ios::sync_with_stdio(false);
//	cin.tie(0);

    int _; scanf("%d",&_);
    while(_--)
    {
    
    
        scanf("%lld%lld",&g,&l);
        if(l%g!=0) {
    
     puts("0"); continue; }
        LL x=l/g;
        map<int,int>mp;
        for(int i=2;i<=x/i;i++)
            if(x%i==0)
                while(x%i==0) x/=i,mp[i]++;
        if(x>1) mp[x]++;
        LL ans=1;
        for(auto x:mp) ans*=6*x.Y;
        printf("%lld\n",ans);
    }



	return 0;
}
/*

*/