Educational Codeforces Round 58 (Rated for Div. 2) D.GCD Counting

D. GCD Counting
time limit per test4.5 seconds
memory limit per test256 megabytes
inputstandard input
outputstandard output
You are given a tree consisting of n vertices. A number is written on each vertex; the number on vertex i is equal to ai.

Let's denote the function g(x,y) as the greatest common divisor of the numbers written on the vertices belonging to the simple path from vertex x to vertex y (including these two vertices). Also let's denote dist(x,y) as the number of vertices on the simple path between vertices x and y, including the endpoints. dist(x,x)=1 for every vertex x.

Your task is calculate the maximum value of dist(x,y) among such pairs of vertices that g(x,y)>1.

Input
The first line contains one integer n — the number of vertices (1≤n≤2⋅105).

The second line contains n integers a1, a2, ..., an (1≤ai≤2⋅105) — the numbers written on vertices.

Then n−1 lines follow, each containing two integers x and y (1≤x,y≤n,x≠y) denoting an edge connecting vertex x with vertex y. It is guaranteed that these edges form a tree.

Output
If there is no pair of vertices x,y such that g(x,y)>1, print 0. Otherwise print the maximum value of dist(x,y) among such pairs.

Examples
inputCopy
3
2 3 4
1 2
2 3
outputCopy
1
inputCopy
3
2 3 4
1 3
2 3
outputCopy
2
inputCopy
3
1 1 1
1 2
2 3
outputCopy
0

题意：

函数g（x，y）表示为：写在属于从顶点x到顶点y（包括这两个顶点）的简单路径的顶点上的数字的最大公约数。

另外，让我们将dist（x，y）表示为顶点x和y之间的简单路径上的顶点数，包括端点。

每个顶点x的dist（x，x）= 1。

任务是计算g（x，y）> 1的顶点对中dist（x，y）的最大值。

如果没有顶点对x，y使得g（x，y）> 1，则打印0.否则在这些对中打印dist（x，y）的最大值。

分析：

将每个gcd分解素因子放入对应vector

相当于对每个素因子建了一棵树

遍历每个素因子的树，寻找直径，更新答案

#include
using namespace std;
#define ll __int64
const int maxn=2e5+10;
int n,a[maxn],vis[maxn],vis1[maxn],ans=1,flag=1,md,mp,p[maxn];
struct edge
{
   int u,v;
   edge(int uu,int vv):u(uu),v(vv){
   }
};
vectore[maxn];
vector v[maxn];
void dfs(int p,int d,int *vis)
{
	if(vis[p])return;
	vis[p]=1;
	if(d>md)md=d,mp=p;
	for(int i=0;i1&&(flag=0);//短路判均<=1 
	} 
	if(flag)
	{
		puts("0");
		return 0; 
	}
	for(int i=1;i1)
		{
		  for(int j=1;j*j<=k;++j)//j从1开始是为了判k 
		  {
		  	if(k%j==0)
		  	{
		  	if(!p[j]&&j!=1)e[j].push_back(edge(u,v));
		  	if(j*j!=k&&!p[k/j])e[k/j].push_back(edge(u,v));
		    }
		  } 
		} 
	}
	for(int i=2;i p;
			for(int j=0;j