HDU5599GTW likes tree

题目
题意：给出一棵点有权的树，求 ∏i,jgcd(val[i],....val[j]) \prod_{i,j}gcd(val[i],....val[j]),就是求所有路径gcd的乘积,保证 1<=val[i]<=105 1

分析：一个简单的思路就是暴力树dp，开一个vector记录每个点向下连出去的gcd的情况，通过前缀积的做法可以做到 O(n∗fac) O(n* fac),然而比较难写，而且空间未必能开得下。
另外一个做法就是点分治，但是复杂度并不优秀。
考虑每个质因子对答案的贡献，由于 val[i]<=105 val[i]一个数不同的质因数不会超过6个.从大到小枚举质因子的幂次，由于我们只关心每个连通块的大小，只需要用并查集维护即可。

#include<bits/stdc++.h>
using namespace std;
const int Maxn=200020,M=1e9+7,N=100020;
typedef pair<int,int>pi;
int isp[N],cnt_pri,pri[N];
vector<int>yinzi[N];
vector<int>G[Maxn];
vector<pi>pro[N];
int n,tl;
int is[Maxn],id[Maxn],ct[22];
int f[Maxn],sz[Maxn],in[Maxn];
int find(int x)
{int a;for(a=x;f[a]!=a;a=f[a]);for(;f[x]!=a;){int t=f[x];f[x]=a;x=t;}return a;
}
int powmod(int x,int y)
{int ret=1;while(y){if(y&1)ret=1LL*ret*x%M;y>>=1;x=1LL*x*x%M;}return ret;
}
int solve()
{int ret=1;for(int it=0;it<cnt_pri;it++){int cur=0;int p=pri[it];if(pro[p].empty())continue;tl++;for(int i=0;i<pro[p].size();i++){int u=pro[p][i].second,w=pro[p][i].first;if(is[u])continue;cur=(cur+w)%(M-1);sz[u]=1;f[u]=u;in[u]=tl;for(int j=0;j<G[u].size();j++){int v=G[u][j];if(in[v]<tl)continue;int uu=find(u),vv=find(v);cur+=1LL*sz[uu]*sz[vv]%(M-1)*w%(M-1);if(cur>=M-1)cur-=M-1;if(sz[uu]>sz[vv]){f[vv]=uu;sz[uu]+=sz[vv];}else{f[uu]=vv;sz[vv]+=sz[uu];}}}ret=1LL*ret*powmod(p,cur)%M;}return ret;
}
int main()
{for(int i=2;i<N;i++){if(!isp[i]){for(int j=i;j<N;j+=i){yinzi[j].push_back(i);isp[j]=1;}pri[cnt_pri++]=i;}}int _;scanf("%d",&_);while(_--){scanf("%d",&n);for(int i=1;i<=n;i++){G[i].clear();is[i]=0;}for(int i=1;i<n;i++){int u,v;scanf("%d%d",&u,&v);G[u].push_back(v);G[v].push_back(u);}for(int i=0;i<cnt_pri;i++)pro[pri[i]].clear();for(int i=1;i<=n;i++){int x;scanf("%d",&x);for(int j=0;j<yinzi[x].size();j++){int cnt=0,tp=x,p=yinzi[x][j];while(tp%p==0)cnt++,tp/=p;pro[p].push_back(pi(cnt,i));}}for(int i=0;i<cnt_pri;i++)sort(pro[pri[i]].begin(),pro[pri[i]].end(),greater<pi>());int ans1=solve();int m;scanf("%d",&m);for(int i=0;i<m;i++){int x;scanf("%d",&x);is[x]=1;}int ans2=solve();ans1=1LL*ans1*powmod(ans2,M-2)%M;printf("%d\n",ans1);}
}

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