Codeforces Round #430 C. Ilya And The Tree

Ilya is very fond of graphs, especially trees. During his last trip to the forest Ilya found a very interesting tree rooted at vertex 1. There is an integer number written on each vertex of the tree; the number written on vertex i is equal to ai.

Ilya believes that the beauty of the vertex x is the greatest common divisor of all numbers written on the vertices on the path from the root to x, including this vertex itself. In addition, Ilya can change the number in one arbitrary vertex to 0 or leave all vertices unchanged. Now for each vertex Ilya wants to know the maximum possible beauty it can have.

For each vertex the answer must be considered independently.

The beauty of the root equals to number written on it.
Input

First line contains one integer number n — the number of vertices in tree (1 ≤ n ≤ 2·105).

Next line contains n integer numbers ai (1 ≤ i ≤ n, 1 ≤ ai ≤ 2·105).

Each of next n - 1 lines contains two integer numbers x and y (1 ≤ x, y ≤ n, x ≠ y), which means that there is an edge (x, y) in the tree.
Output

Output n numbers separated by spaces, where i-th number equals to maximum possible beauty of vertex i.
Examples
Input

2
6 2
1 2

Output

6 6

Input

3
6 2 3
1 2
1 3

Output

6 6 6

Input

1
10

Output
10

题目大意：有一颗树，每一个节点有一个权值，每一个点的美丽度是该点到根节点1的所有点的权值的gcd，在求一个点时，可以去掉路径上任意一个点的权值（不计入gcd中），求每一个点的美丽度最大值

解题报告：我不知道正解是什么，我的暴力水过了，大概思想如下，一个点不选只会影响到子树，然后我们就枚举删除每一个点，然后去更新其子树内的答案，设为f[i],这里明显可以剪枝，如果当前gcd==1,直接返回，记son[x]为x的子树内f的最小值，因为gcd满足递减，所以如果gcd<=son[x] 可以直接返回，然后就跑得过了

#include <algorithm>
#include <iostream>
#include <cstdlib>
#include <cstring>
#include <cstdio>
#include <cmath>
#define RG register
#define il inline
#define iter iterator
#define Max(a,b) ((a)>(b)?(a):(b))
#define Min(a,b) ((a)<(b)?(a):(b))
using namespace std;
const int N=2e5+5;
int n,num=0,head[N],to[N<<1],nxt[N<<1],a[N],f[N],son[N];
void link(int x,int y){
	nxt[++num]=head[x];to[num]=y;head[x]=num;
}
int gcd(int x,int y){
	return x%y?gcd(y,x%y):y;
}
il void updata(int x,int last,int g){
	int u;
	if(g==1)return ;
	if(g<=son[x])return ;
	if(g>f[x])f[x]=g;son[x]=f[x];
	for(RG int i=head[x];i;i=nxt[i]){
		u=to[i];if(u==last)continue;
		updata(u,x,gcd(g,a[u]));
		son[x]=Min(son[x],son[u]);
	}
}
il void dfs(int x,int last,int g){
	int u;
	if(g<=son[x])return ;
	if(g>f[x])f[x]=g;
	for(RG int i=head[x];i;i=nxt[i]){
		u=to[i];if(u==last)continue;
		updata(u,x,g);dfs(u,x,gcd(g,a[u]));
	}
}
void work()
{
	int x,y;
	scanf("%d",&n);
	for(int i=1;i<=n;i++)scanf("%d",&a[i]),f[i]=1;
	for(int i=1;i<n;i++){
		scanf("%d%d",&x,&y);
		link(x,y);link(y,x);
 	}
	dfs(1,1,a[1]);
	for(int i=head[1];i;i=nxt[i]){
		updata(to[i],1,a[to[i]]);
	}
	for(int i=1;i<=n;i++)printf("%d ",f[i]);
}

int main()
{
	work();
	return 0;
}