【题解】CF1101D：GCD Counting

原题传送门
发现要是两个数 $gcd>1$，其实只需要关注他俩是否有质数公因数
对于每个数暴力分解质因数，树形dp
$dp_{u,i}$表示点 $u$为根，第 $i$个质因数的最长链

• $ans=max(dp_{u,i}+dp_{v,j})(fa_v=u,prime_{u,i}=prime_{v,j})$
• $dp_{u,i}=max(dp_{v,j}+1)条件同上$

Code：

#include <bits/stdc++.h>
#define maxn 200010
using namespace std;
vector <int> prime[maxn], dp[maxn];
struct Edge{
	int to, next;
}edge[maxn << 1];
int num, head[maxn], n, ans;

inline int read(){
	int s = 0, w = 1;
	char c = getchar();
	for (; !isdigit(c); c = getchar()) if (c == '-') w = -1;
	for (; isdigit(c); c = getchar()) s = (s << 1) + (s << 3) + (c ^ 48);
	return s * w;
}

void addedge(int x, int y){ edge[++num] = (Edge){y, head[x]}, head[x] = num; }

void dfs(int u, int pre){
	for (int i = head[u]; i; i = edge[i].next){
		int v = edge[i].to;
		if (v != pre){
			dfs(v, u);
			for (int j = 0; j < prime[u].size(); ++j)
				for (int k = 0; k < prime[v].size(); ++k)
					if (prime[u][j] == prime[v][k])
						ans = max(ans, dp[u][j] + dp[v][k]), dp[u][j] = max(dp[u][j], dp[v][k] + 1);
		}
	}
}

int main(){
	n = read();
	for (int i = 1; i <= n; ++i){
		int x = read();
		if (x > 1) ans = 1;
		for (int j = 2; j * j <= x; ++j)
			if (x % j == 0){
				prime[i].push_back(j);
				dp[i].push_back(1);
				while (x % j == 0) x /= j;
			}
		if (x > 1) prime[i].push_back(x), dp[i].push_back(1);
	}
	for (int i = 1; i < n; ++i){
		int x = read(), y = read();
		addedge(x, y), addedge(y, x);
	}
	dfs(1, 0);
	printf("%d\n", ans);
	return 0;
}