合约数

 

借鉴了巨巨的代码，学会看数据量。

 题解：预处理得到每个Val的所有合约数，因为最大的Val才10000。然后在DFS的过程中，对每个节点u，先减去已经存在的u的合约数个数，当回溯到这个节点u的时候，加上u的合约数的个数（子树里面的合约数）。

#pragma warning(disable:4996)
#include<vector>
#include<cstdio>
#include<cstring>
#include<iostream>
#include<algorithm>
#define ll long long 
using namespace std;

const int maxn = 20005;
const int mod = 1000000007;

vector<int> Tree[maxn], H[maxn];

ll ans, F[maxn];
int n, p, kase, cnt[maxn], val[maxn];
bool prime[maxn], use[maxn];

void Inite() {

    memset(prime, 0, sizeof(prime));
    prime[0] = prime[1] = true;

    for (int i = 2; i <= 100; i++) if (!prime[i]) {
        for (int j = i * i; j <= 10000; j += i) {
            prime[j] = true;
        }
    }

    for (int i = 4; i <= 10000; i++) if (prime[i]) {
        for (int j = 4; j <= i; j++) if (prime[j] && (i % j == 0)) {
            H[i].push_back(j);
        }
    }
}

void DFS(int u) {
    for (int i = 0; i < H[val[u]].size(); i++) F[u] -= cnt[H[val[u]][i]];
    cnt[val[u]]++;
    for (int i = 0; i < Tree[u].size(); i++) {
        int v = Tree[u][i];
        if (!use[v]) {
            use[v] = 1;
            DFS(v);
        }
    }
    for (int i = 0; i < H[val[u]].size(); i++) F[u] += cnt[H[val[u]][i]];
    ans += 1LL * u * F[u];
    ans %= mod;
}

int main()
{
    Inite();
    scanf("%d", &kase);
    while (kase--) {
        scanf("%d%d", &n, &p);

        memset(F, 0, sizeof(F));
        memset(cnt, 0, sizeof(cnt));
        memset(use, 0, sizeof(use)); 

        use[p] = 1;
        for (int i = 1; i <= n; i++) Tree[i].clear();

        for (int i = 2; i <= n; i++) {
            int u, v;
            scanf("%d%d", &u, &v);
            Tree[u].push_back(v);
            Tree[v].push_back(u);
        }
        for (int i = 1; i <= n; i++) {
            scanf("%d", val + i);
        }
        
        ans = 0;
        DFS(p);
        printf("%lld\n", ans);
    }
    return 0;
}