CF1101D GCD Counting

题目地址：CF1101D GCD Counting

zz的我比赛时以为是树剖或者点分治然后果断放弃了

这道题不能顺着做，而应该从答案入手反着想

由于一个数的质因子实在太少了，因此首先找到每个点的点权的所有质因子

进行一次树形dp，每次更新暴力枚举所有质因子即可

代码：

#include <bits/stdc++.h>
using namespace std;
const int N = 200006;
int n, ans = 1;
vector<int> p[N], c[N], e[N];
bool flag = 1, v[N];

inline void update(int x, int y) {
    for (unsigned int i = 0; i < p[x].size(); i++)
        for (unsigned int j = 0; j < p[y].size(); j++)
            if (p[x][i] == p[y][j]) {
                ans = max(ans, c[x][i] + c[y][j]);
                c[x][i] = max(c[x][i], c[y][j] + 1);
            }
}

void dp(int x) {
    v[x] = 1;
    for (unsigned int i = 0; i < e[x].size(); i++) {
        int y = e[x][i];
        if (v[y]) continue;
        dp(y);
        update(x, y);
    }
}

inline void divide(int x, int t) {
    for (int i = 2; i * i <= x; i++)
        if (x % i == 0) {
            p[t].push_back(i);
            c[t].push_back(1);
            while (x % i == 0) x /= i;
        }
    if (x > 1) {
        p[t].push_back(x);
        c[t].push_back(1);
    }
}

int main() {
    cin >> n;
    for (int i = 1; i <= n; i++) {
        int x;
        scanf("%d", &x);
        divide(x, i);
        if (x != 1) flag = 0;
    }
    if (flag) {
        puts("0");
        return 0;
    }
    for (int i = 1; i < n; i++) {
        int x, y;
        scanf("%d %d", &x, &y);
        e[x].push_back(y);
        e[y].push_back(x);
    }
    dp(1);
    cout << ans << endl;
    return 0;
}