codeforces1101D GCD Counting 【树形DP】

题目分析：

蛮简单的一道题，对于每个数拆质因子，对于每个质因子找出最长链，在每个地方枚举一下拼接

代码：

 #include<bits/stdc++.h>
 using namespace std;

 const int maxn = ;

 int n,a[maxn],prime[maxn],flag[maxn],minn[maxn],num,ans;
 vector <int> g[maxn];
 vector <pair<int,int> > mp[maxn];

 vector<int> cl[maxn];

 void getprime(int N){
     for(int i=;i<=N;i++){
     if(!flag[i]){prime[++num] = i,minn[i] = i;}
     for(int j=;j<=num&&i*prime[j]<=N;j++){
         flag[i*prime[j]] = ;
         minn[i*prime[j]] = prime[j];
         if(i%prime[j] == ) break;
     }
     }
 }

 void read(){
     scanf("%d",&n);
     for(int i=;i<=n;i++) scanf("%d",&a[i]);
     for(int i=;i<n;i++){
     int u,v; scanf("%d%d",&u,&v);
     g[u].push_back(v); g[v].push_back(u);
     }
 }

 void dp(int now,int fa){
     for(int i=;i<g[now].size();i++){
     if(g[now][i] == fa) continue;
     dp(g[now][i],now);
     }
     int p = a[now];
     while(p != ){cl[minn[p]].clear(); p /= minn[p]; }
     for(int i=;i<g[now].size();i++){
     if(g[now][i] == fa) continue;
     for(int j=;j<mp[g[now][i]].size();j++){
         if(a[now] % mp[g[now][i]][j].first == ){
         cl[mp[g[now][i]][j].first].push_back(mp[g[now][i]][j].second);
         }
     }
     }
     p = a[now];
     while(p != ){
     int z = minn[p]; while(p%z == ) p /= z;
     int maxx = ,sec = ;
     for(int i=;i<cl[z].size();i++){
         if(cl[z][i] >= maxx) sec = maxx,maxx = cl[z][i];
         else if(cl[z][i] > sec) sec = cl[z][i];
     }
     mp[now].push_back(make_pair(z,maxx+));
     ans = max(ans,maxx+sec+);
     }
 }

 void work(){
     getprime();
     dp(,);
     printf("%d\n",ans);
 }

 int main(){
     read();
     work();
     return ;
 }