题目链接：http://acm.fzu.edu.cn/problem.php?pid=2277
Problem Description
Problem DescriptionThere is a rooted tree with n nodes, number from 1-n. Root’s number is 1.Each node has a value ai.
Initially all the node’s value is 0.
We have q operations. There are two kinds of operations.
1 v x k : a[v]+=x , a[v’]+=x-k (v’ is child of v) , a[v’’]+=x-2*k (v’’ is child of v’) and so on.
2 v : Output a[v] mod 1000000007(10^9 + 7).
Input
First line contains an integer T (1 ≤ T ≤ 3), represents there are T test cases.
In each test case:
The first line contains a number n.
The second line contains n-1 number, p2,p3,…,pn . pi is the father of i.
The third line contains a number q.
Next q lines, each line contains an operation. (“1 v x k” or “2 v”)
1?≤?n?≤?3*10^5
1?≤?pi?<?i
1?≤?q?≤?3*10^5
1?≤?v?≤?n; 0?≤?x?<?10^9?+?7; 0?≤?k?<?10^9?+?7
Output
For each operation 2, outputs the answer.
Sample Input
1 3 1 1 3 1 1 2 1 2 1 2 2
Sample Output
2 1
Source
第八届福建省大学生程序设计竞赛-重现赛（感谢承办方厦门理工学院）
题意：略
题目分析：略
1 #include<cstdio>
2 #include<iostream>
3 #include<cstring>
4 #include<algorithm>
5 #include<vector>
6 #define Pii pair<int,int>
7 #define fi first
8 #define se second
9 #define mp make_pair
10 #define pb push_back
11 #define maxn 300100
12 #define LL long long
13 #define lson l,m,rt<<1
14 #define rson m+1,r,rt<<1|1
15 #define Mod 1000000007
16 using namespace std;
17 vector<int> G[maxn];
18 int n,k,x,ti=0;
19 LL tot[maxn];
20 LL C[maxn],add[maxn],Add[maxn*4];
21 int l[maxn],r[maxn];
22 LL deep[maxn];
23 inline int lowbit(int x){return (x&-x);}
24 inline void update(int x,LL y,LL t){
25 while(x<=n){
26 C[x]+=y;
27 tot[x]+=t;
28 x+=lowbit(x);
29 }
30 }
31 LL query(int x){
32 LL ans=0;
33 while (x>0){
34 ans+=C[x];
35 x-=lowbit(x);
36 }
37 return ans;
38 }
39 LL Query(int x){
40 LL ans=0;
41 while (x>0){
42 ans+=tot[x];
43 x-=lowbit(x);
44 }
45 return ans;
46 }
47 void dfs(int x,int fa,int dep){
48 l[x]=++ti;
49 deep[l[x]]=dep;
50 for(int i=0;i<G[x].size();i++){
51 if(G[x][i]==fa) continue;
52 dfs(G[x][i],x,dep+1);
53 }
54 r[x]=ti;
55 }
56 int main(){
57 int u,Q,v,T,opt;
58 cin>>T;
59 while(T--){
60 cin>>n;
61 memset(C,0,sizeof(C));
62 memset(tot,0,sizeof(tot));
63 for(int i=2;i<=n;i++){
64 scanf("%d",&u);
65 G[u].pb(i);
66 }
67 ti=0;
68 dfs(1,-1,1);
69 LL res;
70 cin>>Q;
71 for(int i=1;i<=Q;i++){
72 scanf("%d",&opt);
73 if(opt==1){
74 scanf("%d%d%d",&v,&x,&k);
75 update(l[v],x+(LL)k*deep[l[v]],k);
76 update(r[v]+1,-x-(LL)k*deep[l[v]],-k);
77 }
78 else{
79 scanf("%d",&v);
80 res=query(l[v])-Query(l[v])*deep[l[v]];
81 while(res<0) res+=Mod;
82 res%=Mod;
83 printf("%I64d\n",res);
84 }
85 }
86 for(int i=1;i<=n;i++) G[i].clear();
87 }
88 return 0;
89 }