AtCoder Beginner Contest 351
翻译中…
Problem Statement
翻译中…
You are given a rooted tree with $N$ vertices numbered $1$ to $N$.
Vertex $1$ is the root, and the parent of vertex $i$ $(2 \leq i \leq N)$ is vertex $p_i$ $(p_i < i)$.
Additionally, there is a sequence $A = (A_1, A_2, \dots, A_N)$.
翻译中…
The hash value of the rooted tree is calculated as follows:
翻译中…
- Define
$f(n)$$(1 \leq n \leq N)$as follows in the order$n = N, N-1, \dots, 2, 1$. - If vertex
$n$is a leaf,$f(n) = A_n$.- If vertex
$n$is not a leaf,$\displaystyle f(n) = A_n + \prod_{c \in C(n)} f(c)$, where$C(n)$is the set of children of$n$.
- If vertex
- The hash value of the rooted tree is
$f(1) \bmod{998244353}$.
翻译中…
Process $Q$ queries in the order they are given.
Each query provides $v$ and $x$, so update $A_v$ to $x$ and then compute the hash value of the rooted tree.
翻译中…
Constraints
翻译中…
$2 \leq N \leq 2 \times 10^5$$1 \leq Q \leq 2 \times 10^5$$1 \leq p_i < i$$0 \leq A_i < 998244353$$1 \leq v \leq N$$0 \leq x < 998244353$- All input values are integers.
翻译中…
Input
翻译中…
The input is given from Standard Input in the following format, where $\mathrm{query}_i$ represents the $i$-th query:
翻译中…
N Q
p_2 p_3 \dots p_N
A_1 A_2 \dots A_N
\mathrm{query}_1
\mathrm{query}_2
\vdots
\mathrm{query}_Q
翻译中…
Each query is given in the following format:
翻译中…
v x
翻译中…
Output
翻译中…
Print $Q$ lines. The $i$-th line should contain the answer to the $i$-th query.
翻译中…
样例
Input #1
3 2
1 1
3 5 1
3 4
2 1
Output #1
23
7
Initially, $A = (3, 5, 1)$.
The first query is processed as follows:
- Update $A_3$ to $4$. Now $A = (3, 5, 4)$.
- The hash value of the rooted tree is calculated as follows to be $23$, which should be printed.
- Vertex $3$ has no children. Thus, $f(3) = 4$.
- Vertex $2$ has no children. Thus, $f(2) = 5$.
- Vertex $1$ has children $2$ and $3$. Thus, $f(1) = 3 + 5 \times 4 = 23$.
- $f(1) \bmod{998244353} = 23$ is the hash value of the rooted tree.
The second query is processed as follows:
- Update $A_2$ to $1$. Now $A = (3, 1, 4)$.
- The hash value of the rooted tree is calculated as follows to be $7$:
- Vertex $3$ has no children. Thus, $f(3) = 4$.
- Vertex $2$ has no children. Thus, $f(2) = 1$.
- Vertex $1$ has children $2$ and $3$. Thus, $f(1) = 3 + 1 \times 4 = 7$.
- $f(1) \bmod{998244353} = 7$ is the hash value of the rooted tree.
Input #2
5 4
1 1 2 2
2 5 4 4 1
3 3
5 0
4 5
5 2
Output #2
29
17
17
47
Input #3
10 10
1 2 1 2 5 6 3 5 1
766294629 440423913 59187619 725560240 585990756 965580535 623321125 550925213 122410708 549392044
1 21524934
9 529970099
6 757265587
8 219853537
5 687675301
5 844033519
8 780395611
2 285523485
6 13801766
3 487663184
Output #3
876873846
952166813
626349486
341294449
466546009
331098453
469507939
414882732
86695436
199797684
题目信息
题面与样例版权归 AtCoder 及出题人所有;中文翻译由 AI 生成(CI 管线),如有出入请以原文为准。