cf396C. On Changing Tree

题解 操作一可以转化成 x+dep[i]*k-dep[j]*k  j是i的子树节点 然后dfs序以后用两个树状数组维护即可

#include <algorithm>
#include <iostream>
#include <cstring>
#include <cstdio>
#include <vector>
#include <stack>
#include <queue>
#include <cmath>
#include <set>
#include <map>
#define mp make_pair
#define pb push_back
#define pii pair<int,int>
#define link(x) for(edge *j=h[x];j;j=j->next)
#define inc(i,l,r) for(int i=l;i<=r;i++)
#define dec(i,r,l) for(int i=r;i>=l;i--)
const int MAXN=3e5+10;
const double eps=1e-8;
#define ll long long
const int mod=1e9+7;
using namespace std;
struct edge{int t;edge*next;}e[MAXN],*h[MAXN],*o=e;
void add(int x,int y){o->t=y;o->next=h[x];h[x]=o++;}
ll read(){
    ll x=0,f=1;char ch=getchar();
    while(!isdigit(ch)){if(ch=='-')f=-1;ch=getchar();}
    while(isdigit(ch))x=x*10+ch-'0',ch=getchar();
    return x*f;
}
ll sum1[MAXN],sum2[MAXN];
int p[MAXN],cnt,dep[MAXN],num[MAXN];
void dfs(int x,int fa,int deep){
	p[x]=++cnt;dep[x]=deep+1;num[x]=1;
	link(x){
		if(j->t!=fa){
			dfs(j->t,x,deep+1);
			num[x]+=num[j->t];
		}
	}
}
int n;
int get_id(int x){return x&(-x);}
void update1(int x,ll k){
	for(int i=x;i<=n+1;i+=get_id(i))sum1[i]+=k,sum1[i]%=mod;
}
void update2(int x,ll k){
	for(int i=x;i<=n+1;i+=get_id(i))sum2[i]+=k,sum2[i]%=mod;
}
ll Sum1(int x){
	ll ans=0;
	for(int i=x;i>0;i-=get_id(i))ans+=sum1[i],ans%=mod;
	return ans;
}
ll Sum2(int x){
	ll ans=0;
	for(int i=x;i>0;i-=get_id(i))ans+=sum2[i],ans%=mod;
	return ans;
}
int main(){
	n=read();
	int u,v,op;ll x,k;
	inc(i,2,n)u=read(),add(u,i);
	dfs(1,0,0);
	int q=read();
	while(q--){
		op=read();
		if(op==1){
			v=read();x=read();k=read();
			ll t1=(x+(1ll*dep[v]*k%mod))%mod;
			//cout<<t1<<endl;
			update1(p[v],t1);
			update1(p[v]+num[v],(mod-t1));
			update2(p[v],k);
			update2(p[v]+num[v],mod-k);
		}
		else{
			v=read();
			ll ans=Sum1(p[v]);
			//cout<<ans<<"==="<<endl;
			ll ans2=Sum2(p[v]);
			ans2=ans2*dep[v]%mod;
			ans=(ans-ans2+mod)%mod;
			printf("%lld
",ans);
		}
	}
	return 0;
}

C. On Changing Tree

time limit per test

2 seconds

memory limit per test

256 megabytes

input

standard input

output

standard output

You are given a rooted tree consisting of n vertices numbered from 1 to n. The root of the tree is a vertex number 1.

Initially all vertices contain number 0. Then come q queries, each query has one of the two types:

• The format of the query: 1 v x k. In response to the query, you need to add to the number at vertex v number x; to the numbers at the descendants of vertex v at distance 1, add x - k; and so on, to the numbers written in the descendants of vertex v at distance i, you need to add x - (i·k). The distance between two vertices is the number of edges in the shortest path between these vertices.
• The format of the query: 2 v. In reply to the query you should print the number written in vertex v modulo 1000000007 (109 + 7).

Process the queries given in the input.

Input

The first line contains integer n (1 ≤ n ≤ 3·105) — the number of vertices in the tree. The second line contains n - 1 integers p2, p3, ... pn(1 ≤ pi < i), where pi is the number of the vertex that is the parent of vertex i in the tree.

The third line contains integer q (1 ≤ q ≤ 3·105) — the number of queries. Next q lines contain the queries, one per line. The first number in the line is type. It represents the type of the query. If type = 1, then next follow space-separated integers v, x, k (1 ≤ v ≤ n; 0 ≤ x < 109 + 7; 0 ≤ k < 109 + 7). If type = 2, then next follows integer v (1 ≤ v ≤ n) — the vertex where you need to find the value of the number.

Output

For each query of the second type print on a single line the number written in the vertex from the query. Print the number modulo 1000000007 (109 + 7).

Examples

input

Copy

3
1 1
3
1 1 2 1
2 1
2 2

output

Copy

2
1

Note

You can read about a rooted tree here: http://en.wikipedia.org/wiki/Tree_(graph_theory).