Educational Codeforces Round 45 (Rated for Div. 2) G - GCD Counting

G - GCD Counting

思路：我猜测了一下gcd的个数不会很多，然后我就用dfs回溯的时候用map暴力合并就好啦。

#include<bits/stdc++.h>
#define LL long long
#define fi first
#define se second
#define mk make_pair
#define pii pair<int,int>
#define piii pair<int, pair<int,int> >

using namespace std;

const int N = 2e5 + 7;
const int M = 10 + 7;
const int inf = 0x3f3f3f3f;
const LL INF = 0x3f3f3f3f3f3f3f3f;
const int mod = 1e9 + 7;
const double eps = 1e-6;

int n, a[N];
LL cnt[N];
vector<int> edge[N];
map<int, LL> mp[N];

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

void dfs(int u, int p) {
    mp[u][a[u]]++;
    for(int v : edge[u]) {
        if(v == p) continue;
        dfs(v, u);
        map<int, LL> :: iterator itu, itv;
        for(itu = mp[u].begin(); itu != mp[u].end(); itu++) {
            for(itv = mp[v].begin(); itv != mp[v].end(); itv++) {
                cnt[gcd(itu -> first, itv -> first)] += itu -> second * itv -> second;
            }
        }

        for(itv = mp[v].begin(); itv != mp[v].end(); itv++) {
            mp[u][gcd(a[u], itv -> first)] += itv -> second;
        }
    }
}

int main() {
    scanf("%d", &n);
    for(int i = 1; i <= n; i++) {
        scanf("%d", &a[i]);
        cnt[a[i]]++;
    }

    for(int i = 1; i < n; i++) {
        int u, v; scanf("%d%d", &u, &v);
        edge[u].push_back(v);
        edge[v].push_back(u);
    }

    dfs(1, 0);

    for(int i = 1; i < N; i++) {
        if(cnt[i]) {
            printf("%d %lld\n", i, cnt[i]);
        }
    }
    return 0;
}
/*
*/