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 a_i.

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

For each vertex the answer must be considered independently.

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

給出一棵生成樹，每個節點都有一個值，現在要求出每個節點的美麗值的最大值，美麗值的定義為從根節點到該節點(包含)路徑上所有點的值的gcd，求解每個 點時可以把路徑上某一個點的值變為0（就相當於刪除這個節點的數）。你可以認為每個點美麗值的求解是獨立的（每次刪除的點都不會影響下一次）。

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

Next line contains n integer numbers a_i (1?≤?i?≤?n, 1?≤?a_i?≤?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 n numbers separated by spaces, where i-th number equals to maximum possible beauty of vertex i.

2
6 2
1 2

6 6 

3
6 2 3
1 2
1 3

6 6 6 

1
10

10 

我們用c[i]表示從1～i不變化的gcd值

用set[i]表示已變化的gcd的值的集合

對於u->v有

set[v].insert(c[u])表示將v變化

set[v][]=gcd(set[u][i],a[v])在v之前已變化

但這樣可能存在疑問，每一次的集合都增加一個，而且還要遍歷

這豈不是n^2???

但set去重後數量卻遠遠達不到n，也就√n

因為a[x]的因數總共就不超過√a[x]個，不可能超過這麽多,而a[x]<=20^5

所以可以過

輸出直接輸出集合中最大的數

代碼這裏用了STL中的集合

%%%%yzh巨佬用n^2暴力強行過：

http://www.cnblogs.com/Yuzao/p/7451552.html

果然巨佬還是強無敵啊

 1 #include<iostream>
 2 #include<cstdio>
 3 #include<cstring>
 4 #include<algorithm>
 5 #include<set>
 6 using namespace std;
 7 struct Node
 8 {
 9   int next,to;
10 }edge[400001];
11 int num,head[200001],c[200001],a[200001],n;
12 set<int>Set[200001];
13 void add(int u,int v)
14 {
15   num++;
16   edge[num].next=head[u];
17   head[u]=num;
18   edge[num].to=v;
19 }
20 int gcd(int a,int b)
21 {
22   if (!b) return a;
23   return gcd(b,a%b);
24 }
25 void dfs(int x,int pa)
26 {
27   for (int i=head[x];i;i=edge[i].next)
28     {
29       int v=edge[i].to;
30       if (v==pa) continue;
31       c[v]=gcd(c[x],a[v]);
32       Set[v].insert(c[x]);
33       for (set<int>::iterator i=Set[x].begin();i!=Set[x].end();i++)
34     Set[v].insert(gcd(a[v],*i));
35       dfs(v,x);
36     }
37 }
38 int main()
39 {int i,u,v;
40   cin>>n;
41   for (i=1;i<=n;i++)
42     {
43       scanf("%d",&a[i]);
44     }
45   for (i=1;i<=n-1;i++)
46     {
47       scanf("%d%d",&u,&v);
48       add(u,v);add(v,u);
49     }
50   c[1]=a[1];
51   Set[1].insert(0);
52   dfs(1,0);
53   cout<<a[1]<<‘ ‘;
54   for (i=2;i<n;i++)
55     printf("%d ",*(--Set[i].end()));
56   if (n>1)
57   cout<<*(--Set[n].end())<<endl;
58 }

codeforces 842C Ilya And The Tree