传送门
文章目录
题意：思路：题意：
给三个数的lcmlcmlcm和gcdgcdgcd，求满足条件的三元组组合个数。
思路：
首先lcmmodgcd==0lcm\bmod gcd==0lcmmodgcd==0是有组合的条件，否则输出0。
现在可知lcm(x′,y′,z′)=lcm(x,y,z)gcd(x,y,z)，gcd(x′,y′,z′)=1lcm(x^{'},y^{'},z^{'})=\frac{lcm(x,y,z)}{gcd(x,y,z)}，gcd(x^{'},y^{'},z^{'})=1lcm(x′,y′,z′)=gcd(x,y,z)lcm(x,y,z)，gcd(x′,y′,z′)=1，对aaa分解质因子得到p1u1p2u2...pnunp_1^{u_1}p_2^{u_2}...p_n^{u_n}p1u1p2u2...pnun，假设x′=p1i1p2i2...pnjn,y′=p1j1p2j2...pnjn,z′=p1k1p2k2...pnknx^{'}=p_1^{i_1}p_2^{i_2}...p_n^{j_n},y^{'}=p_1^{j_1}p_2^{j_2}...p_n^{j_n},z^{'}=p_1^{k_1}p_2^{k_2}...p_n^{k_n}x′=p1i1p2i2...pnjn,y′=p1j1p2j2...pnjn,z′=p1k1p2k2...pnkn。那么由于gcd(x′,y′,z′)=1gcd(x^{'},y^{'},z^{'})=1gcd(x′,y′,z′)=1，可知min(i1,j1,k1)=0min(i_1,j_1,k_1)=0min(i1,j1,k1)=0,max(i1,j1,k1)=u1max(i_1,j_1,k_1)=u_1max(i1,j1,k1)=u1，所以我们需要找出来一个位置取000，一个位置取u1u_1u1，其他的位置随意就好了。当前位置的答案即为A32∗u1=6∗u1A_3^2*u_1=6*u_1A32∗u1=6∗u1，那么ans=∑6∗uians=\sum6*u_ians=∑6∗ui。
//#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;}/**/
