Gym 100818B Tree of Almost Clean Money

题意：给你一棵树，每次一个操作，修改书上最多不超过1000个点的值，然后询问u到v路径上的权值和。

分析：树链剖分，后边单点修改区间查询可以用树状数组。

#include<iostream>
#include<string>
#include<algorithm>
#include<cstdlib>
#include<cstdio>
#include<vector>
#include<stack>
#include<queue>
#define INF 0x3f3f3f3f
#define MOD 1000000007
#define N 500005
using namespace std;
typedef long long ll;
ll val[N],f[4*N];
int dep[N],siz[N],fa[N],id[N],son[N],top[N],fid[N];
int T,q,n,num;
vector <int> G[N];
struct Segtree
{
	int l,r;
	ll val;
} tree[4*N];
void dfs1(int u,int f,int d)
{
	dep[u] = d;
	siz[u] = 1;
	son[u] = 0;
	fa[u] = f;
	for(int v : G[u])
	 if(v != f)
	 {
	 	 dfs1(v,u,d+1);
	 	 siz[u] += siz[v];
	 	 if(siz[son[u]] < siz[v]) son[u] = v;
	 }
}
int lowbit(int x)
{
	return (x & (-x));
}
void Insert(int x,ll val)
{
	while(x <= n)
	{
		f[x] += val;
		x += lowbit(x); 
	}
}
ll Find(int x)
{
	ll ans = 0;
	while(x)
	{
		ans += f[x];
		x -= lowbit(x);
	}
	return ans;
}
ll query(int x,int y)
{
	return (Find(y) - Find(x-1));
}
void dfs2(int u,int tp)
{
	top[u] = tp;
	id[u] = ++num;
	fid[id[u]] = u;
	if(son[u]) dfs2(son[u],tp);
	for(int v : G[u])
	 if(v != fa[u] && v != son[u]) dfs2(v,v);
}
ll Yougth(int u,int v)
{
	int tp1 = top[u],tp2 = top[v];
	ll ans = 0;
	while(tp1 != tp2)
	{
		if(dep[tp1] < dep[tp2])
		{
			swap(tp1,tp2);
			swap(u,v);
		}
		ans += query(id[tp1],id[u]);
		u = fa[tp1];
		tp1 = top[u]; 
	}
	if(dep[u] > dep[v]) swap(u,v);
	ans += query(id[u],id[v]);
	return ans;
}
int main()
{
		scanf("%d",&n);
		for(int i = 1;i < n;i++)
		{
			int x,y;
			scanf("%d%d",&x,&y);
			x++,y++;
			G[x].push_back(y);
			G[y].push_back(x);
		}
		num = 0;
		dfs1(1,0,1);
		dfs2(1,1);
		for(int i = 1;i <= n;i++) 
		{
			scanf("%I64d",&val[i]);
			Insert(id[i],val[i]);
		}
		scanf("%d",&q);
		for(int i = 1;i <= q;i++)
		{
			int K,u,v;
			ll x1,y1,A,B,C,D;
			scanf("%d%I64d%I64d%I64d%I64d%I64d%I64d%d%d",&K,&x1,&y1,&A,&B,&C,&D,&u,&v);
			Insert(id[x1+1],y1);
			for(int j = 1;j < K;j++)
			{
				x1 = (A*x1 + B),y1 = (C*y1 + D);
				if(x1 >= n) x1 = x1 % n;
				if(y1 >= MOD) y1 = y1 % MOD;
				Insert(id[x1+1],y1);
			}
			u++,v++;
			printf("%I64d\n",Yougth(u,v));
		}
		return 0;
}