Codeforces Round #264 (Div. 2)E（裸的树询问）

最新推荐文章于 2025-08-01 11:12:48 发布

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最新推荐文章于 2025-08-01 11:12:48 发布

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分类专栏： codeforces 文章标签： ACM codeforces

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Caisa is now at home and his son has a simple task for him.

Given a rooted tree with n vertices, numbered from 1 to n (vertex 1 is the root). Each vertex of the tree has a value. You should answer qqueries. Each query is one of the following:

Format of the query is "1 v". Let's write out the sequence of vertices along the path from the root to vertex v: u1, u2, ..., uk (u1 = 1; uk = v). You need to output such a vertex ui that gcd(value of ui, value of v) > 1 and i < k. If there are several possible vertices ui pick the one with maximum value of i. If there is no such vertex output -1.
Format of the query is "2 v w". You must change the value of vertex v to w.

You are given all the queries, help Caisa to solve the problem.

Input

The first line contains two space-separated integers n, q (1 ≤ n, q ≤ 105).

The second line contains n integers a1, a2, ..., an (1 ≤ ai ≤ 2·106), where ai represent the value of node i.

Each of the next n - 1 lines contains two integers xi and yi (1 ≤ xi, yi ≤ n; xi ≠ yi), denoting the edge of the tree between vertices xiand yi.

Each of the next q lines contains a query in the format that is given above. For each query the following inequalities hold: 1 ≤ v ≤ n and1 ≤ w ≤ 2·106. Note that: there are no more than 50 queries that changes the value of a vertex.

Output

For each query of the first type output the result of the query.

Sample test(s)

input
4 6
10 8 4 3
1 2
2 3
3 4
1 1
1 2
1 3
1 4
2 1 9
1 4

output
-1
1
2
-1
1

题意：两种操作 RT

思路：纯暴力真的能过，也就900多MS，CF好逗，sad~

#include 
#include 
#include 
#include 
using namespace std;

int fa[100010];
int head[100010];
int edge[100010<<1];
int next[100010<<1];
int ww[100010];
int d;
int n;
int q;

void add(int u,int v)
{
    edge[d]=v;
    next[d]=head[u];
    head[u]=d++;
}

void dfs(int u,int pre)
{
    fa[u]=pre;
    for(int i=head[u];i!=-1;i=next[i])if(edge[i]!=pre)dfs(edge[i],u);
}

int gcd(int a,int b)
{
    if(a1){
                maxi=u;
                break;
        }
        u=fa[u];
    }
    return maxi;
}

int main()
{
    while(scanf("%d%d",&n,&q)!=EOF)
    {
        int i;
        int u,v;
        memset(head,-1,sizeof(head));
        d=0;

        for(i=1;i<=n;i++)scanf("%d",&ww[i]);

        for(i=1;i