传送门
文章目录
- 题意:
- 思路:
题意:
给你一颗树,其中根是111,每个点有一个点权,求每个点到根的所有路径的gcdgcdgcd之和。
n≤1e5n\le1e5n≤1e5
思路:
一看到以为是个点分治,让后发现不是任意点对所以显然不能用点分治来做。
后来想了半天也只想到一个暴力,没想到还真是个暴力。
考虑从根到点iii的路径上同的gcdgcdgcd个数,可以证明最多有log2(1012)log_2(10^{12})log2(1012)个,所以每个点都继承一下父节点的值即可,由于需要用mapmapmap维护,所以复杂度O(nlog22(1012))O(nlog^2_2(10^{12}))O(nlog22(1012))。
// Problem: E. Kamil and Making a Stream
// Contest: Codeforces - Codeforces Round #588 (Div. 2)
// URL: https://codeforces.com/contest/1230/problem/E
// Memory Limit: 768 MB
// Time Limit: 4000 ms
//
// Powered by CP Editor (https://cpeditor.org)//#pragma GCC optimize("Ofast,no-stack-protector,unroll-loops,fast-math")
//#pragma GCC target("sse,sse2,sse3,ssse3,sse4.1,sse4.2,avx,avx2,popcnt,tune=native")
//#pragma GCC optimize(2)
#include<cstdio>
#include<iostream>
#include<string>
#include<cstring>
#include<map>
#include<cmath>
#include<cctype>
#include<vector>
#include<set>
#include<queue>
#include<algorithm>
#include<sstream>
#include<ctime>
#include<cstdlib>
#include<random>
#include<cassert>
#define X first
#define Y second
#define L (u<<1)
#define R (u<<1|1)
#define pb push_back
#define mk make_pair
#define Mid ((tr[u].l+tr[u].r)>>1)
#define Len(u) (tr[u].r-tr[u].l+1)
#define random(a,b) ((a)+rand()%((b)-(a)+1))
#define db puts("---")
using namespace std;//void rd_cre() { freopen("d://dp//data.txt","w",stdout); srand(time(NULL)); }
//void rd_ac() { freopen("d://dp//data.txt","r",stdin); freopen("d://dp//AC.txt","w",stdout); }
//void rd_wa() { freopen("d://dp//data.txt","r",stdin); freopen("d://dp//WA.txt","w",stdout); }typedef long long LL;
typedef unsigned long long ULL;
typedef pair<int,int> PII;const int N=1000010,mod=1e9+7,INF=0x3f3f3f3f;
const double eps=1e-6;int n;
vector<int>v[N];
vector<pair<LL,LL>>all[N];
LL a[N],ans;
map<LL,int>mp;void del() {if(ans>=mod) ans-=mod;
}void dfs(int u,int f) {mp.clear();ans+=a[u]; del();for(auto y:all[f]) {ans+=y.Y*__gcd(y.X,a[u])%mod; ans%=mod;mp[__gcd(y.X,a[u])]+=y.Y;}mp[a[u]]++;for(auto x:mp) all[u].pb({x.X,x.Y});for(auto x:v[u]) {if(x==f) continue;dfs(x,u);}
}int main()
{
// ios::sync_with_stdio(false);
// cin.tie(0);cin>>n;for(int i=1;i<=n;i++) scanf("%lld",&a[i]);for(int i=1;i<=n-1;i++) {int a,b; scanf("%d%d",&a,&b);v[a].pb(b); v[b].pb(a);}dfs(1,0);printf("%lld\n",ans%mod);return 0;
}
/**/