Codeforces Round #588 (Div. 2) E. Kamil and Making a Stream(DFS)

链接:

https://codeforces.com/contest/1230/problem/E

题意:

Kamil likes streaming the competitive programming videos. His MeTube channel has recently reached 100 million subscribers. In order to celebrate this, he posted a video with an interesting problem he couldn't solve yet. Can you help him?

You're given a tree — a connected undirected graph consisting of n vertices connected by n−1 edges. The tree is rooted at vertex 1. A vertex u is called an ancestor of v if it lies on the shortest path between the root and v. In particular, a vertex is an ancestor of itself.

Each vertex v is assigned its beauty xv — a non-negative integer not larger than 1012. This allows us to define the beauty of a path. Let u be an ancestor of v. Then we define the beauty f(u,v) as the greatest common divisor of the beauties of all vertices on the shortest path between u and v. Formally, if u=t1,t2,t3,…,tk=v are the vertices on the shortest path between u and v, then f(u,v)=gcd(xt1,xt2,…,xtk). Here, gcd denotes the greatest common divisor of a set of numbers. In particular, f(u,u)=gcd(xu)=xu.

Your task is to find the sum

∑u is an ancestor of vf(u,v).
As the result might be too large, please output it modulo 109+7.

Note that for each y, gcd(0,y)=gcd(y,0)=y. In particular, gcd(0,0)=0.

思路:

暴力题..map记录每个点有多少个gcd的值, 从父节点继承下来.

代码:

#include 
using namespace std;
typedef long long LL;
const int MAXN = 1e5+10;
const int MOD = 1e9+7;

LL a[MAXN], ans = 0;
vector G[MAXN];
unordered_map Mp[MAXN];
int n;

void Dfs(int u, int fa)
{
    for (auto it: Mp[fa])
    {
        LL gcd = __gcd(a[u], it.first);
        Mp[u][gcd] += it.second;
    }
    Mp[u][a[u]]++;
    for (auto it: Mp[u])
        ans = (ans + (it.first*it.second)%MOD)%MOD;
    for (auto x: G[u])
    {
        if (x == fa)
            continue;
        Dfs(x, u);
    }
}

int main()
{
    cin >> n;
    for (int i = 1;i <= n;i++)
        cin >> a[i];
    int u, v;
    for (int i = 1;i < n;i++)
    {
        cin >> u >> v;
        G[u].push_back(v);
        G[v].push_back(u);
    }
    Dfs(1, 0);
    cout << ans << endl;

    return 0;
}