codeforces Round #264(div2) E解题报告

E. Caisa and Tree

time limit per test

10 seconds

memory limit per test

256 megabytes

input

standard input

output

standard output

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 xi and 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 and 1 ≤ 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

Note

gcd(x, y) is greatest common divisor of two integers x and y.

题目大意：

给出一个总节点数量为n的树，每个节点有权值，进行q次操作，每次操作有两种选项：

       1. 询问节点v到root之间的路径上的各个节点，求满足条件 gcd(val[i], val[v]) > 1 的 距离v最近的节点的下标。

       2. 将节点v的值求改为w。

解法：

时间限制为10s，给的好宽松，直接暴力dfs过。

代码：

#include <cstdio>
#include <vector>
#define N_max 123456

using namespace std;

int  n, m, ans;
int  val[N_max], fa[N_max];
vector <int>  G[N_max];

int gcd(int a, int b) {
	return b ? gcd(b, a%b) : a;
}

void build(int v, int fav) {
	fa[v] = fav;

	for (int i = 0; i < G[v].size(); i++)
		if (G[v][i] != fav) {
			fa[G[v][i]] = v;
			build(G[v][i], v);
		}
}

void addedge(int x, int y) {
	G[x].push_back(y);
}

void init() {
	scanf("%d%d", &n, &m);
	for (int i = 1; i <= n; i++)  scanf("%d", &val[i]);

	for (int i = 1; i <= n-1; i++) {
		int x, y;
		
		scanf("%d%d", &x, &y);
		addedge(x, y);
		addedge(y, x);
	}

	build(1, 0);
}

int query(int v) {
	int w=val[v];

	v = fa[v];
	while (v) {
		if (gcd(val[v], w) > 1)
			return v;

		v = fa[v];
	}

	return -1;
}

void solve() {
	for (int i = 1; i <= m; i++) {
		int tmp, v, w;

		scanf("%d", &tmp);

		if (tmp == 1) {
			scanf("%d", &v);
			printf("%d\n", query(v));
		}
		else {
			scanf("%d%d", &v, &w);
			val[v] = w;
		}
	}
}

int main() {
	init();
	solve();
}