CF EDU 1101D GCD Counting 

題意

　　 有一顆樹，每個節點有一個值，問樹上最長鏈的長度，要求鏈上的每個節點的GCD值大於1。

思路

　　由於每個數的質因子很少，題目的資料200000<2*3*5*7*11*13*17=510510。所以每個節點的質因子個數不多。那麼樹形DP的時候直接列舉每種因子即可。

//#pragma GCC optimize(3)
//#pragma comment(linker, "/STACK:102400000,102400000")  //c++
// #pragma GCC diagnostic error "-std=c++11"
// #pragma comment(linker, "/stack:200000000")
 
// #pragma GCC target("sse,sse2,sse3,ssse3,sse4,popcnt,abm,mmx,avx,tune=native")
 
#include <algorithm>
#include  <iterator>
#include  <iostream>
#include   <cstring>
#include   <cstdlib>
#include   <iomanip>
#include    <bitset>
#include    <cctype>
#include    
 
<cstdio>
#include    <string>
#include    <vector>
#include     <stack>
#include     <cmath>
#include     <queue>
#include      <list>
#include       <map>
#include       <set>
#include   <cassert>

using namespace std;
#define lson (l , mid , rt << 1)
#define
 
 rson (mid + 1 , r , rt << 1 | 1)
#define debug(x) cerr << #x << " = " << x << "\n";
#define pb push_back
#define pq priority_queue
 
 

typedef long long ll;
typedef unsigned long long ull;
//typedef __int128 bll;
typedef pair<ll ,ll > pll;
typedef pair<int ,int > pii;
typedef pair<int,pii> p3;
 
//priority_queue<int> q;//這是一個大根堆q
//priority_queue<int,vector<int>,greater<int> >q;//這是一個小根堆q
#define fi first
#define se second
//#define endl '\n'
 
#define OKC ios::sync_with_stdio(false);cin.tie(0)
#define FT(A,B,C) for(int A=B;A <= C;++A)  //用來壓行
#define REP(i , j , k)  for(int i = j ; i <  k ; ++i)
#define max3(a,b,c) max(max(a,b), c);
#define min3(a,b,c) min(min(a,b), c);
//priority_queue<int ,vector<int>, greater<int> >que;
 
const ll mos = 0x7FFFFFFF;  //2147483647
const ll nmos = 0x80000000;  //-2147483648
const int inf = 0x3f3f3f3f;
const ll inff = 0x3f3f3f3f3f3f3f3f; //18
// const int mod = 998244353;
const double esp = 1e-8;
const double PI=acos(-1.0);
const double PHI=0.61803399;    //黃金分割點
const double tPHI=0.38196601;
 
 
template<typename T>
inline T read(T&x){
    x=0;int f=0;char ch=getchar();
    while (ch<'0'||ch>'9') f|=(ch=='-'),ch=getchar();
    while (ch>='0'&&ch<='9') x=x*10+ch-'0',ch=getchar();
    return x=f?-x:x;
} 
/*-----------------------showtime----------------------*/

            const int maxn = 2e5+9;
            int a[maxn];

            vector<int>mp[maxn];
            int vis[maxn];
            int dp[maxn];
            int ans = 0;
            vector<int>p[maxn],t[maxn];

            void dfs(int u, int fa){
                for(int i=0; i<mp[u].size(); i++){
                    int v = mp[u][i];
                    if(v == fa)continue;
                    dfs(v,u);

                    for(int j=0; j<p[u].size(); j++){
                        for(int k=0; k<p[v].size(); k++){
                            if(p[u][j] == p[v][k]){
                                ans = max(ans, t[u][j] + t[v][k]);
                                t[u][j] = max(t[u][j], t[v][k] + 1);
                            }
                        }
                    }
                }
                if(a[u] > 1) ans = max(ans, 1);
            }

int main(){
            int n;  scanf("%d", &n);
            int flag = 1;
            for(int i=1; i<=n; i++) {
                scanf("%d", &a[i]);
                int x = a[i];
                for(ll j=2; j*j <=x; j++){
                    if(x%j == 0){
                        p[i].pb(j);
                        t[i].pb(1);
                        while(x%j==0) x/=j;
                    }
                }
                if(x > 1){
                    p[i].pb(x);
                    t[i].pb(1);
                }
                if(a[i] > 1) flag = 0;
            }

            if(flag) {
                puts("0");
                return 0;
            }
            for(int i=1; i<n; i++){
                int u,v;
                scanf("%d%d", &u, &v);
                mp[u].pb(v);
                mp[v].pb(u);
            }

            dfs(1,-1);
            printf("%d\n", ans);
            return 0;
}