Codeforces 842C Ilya And The Tree【Dfs】

C. Ilya And The Tree

time limit per test
2 seconds

memory limit per test
256 megabytes

input
standard input

output
standard output

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：

求从根节点1到每一个节点x路径上的gcd最大值，我们可以用一次机会使得路径上的某个数变成0.

思路：

爆搜+set去重即可。因为每个数的因子数最多为logn个，那么对应一条路径上的gcd的值也不会多。

Ac代码：

#include<stdio.h>
#include<string.h>
#include<vector>
#include<set>
using namespace std;
vector<int>mp[250000];
int ans[250000];
int a[250000];

int gcd(int x,int y)
{
return y==0?x:gcd(y,x%y);
}
void Dfs(int u,int from,set<int>now,int Val)
{
for(int i=0;i<mp[u].size();i++)
{
int v=mp[u][i];
if(v==from)continue;
set<int>nex;
nex.insert(Val);
set<int>::iterator it;
for(it=now.begin();it!=now.end();it++)
{
nex.insert(gcd(*it,a[v]));
}
ans[v]=max(ans[v],*nex.rbegin());
Dfs(v,u,nex,gcd(Val,a[v]));
}
}
int main()
{
int n;
while(~scanf("%d",&n))
{
for(int i=1;i<=n;i++)mp[i].clear();
for(int i=1;i<=n;i++)scanf("%d",&a[i]);
for(int i=1;i<=n-1;i++)
{
int x,y;scanf("%d%d",&x,&y);
mp[x].push_back(y);
mp[y].push_back(x);
}
ans[1]=a[1];
set<int>temp;temp.clear();
temp.insert(a[1]);
temp.insert(0);
Dfs(1,-1,temp,a[1]);
for(int i=1;i<=n;i++)printf("%d ",ans[i]);
printf("\n");
}
}