Bracket Insertion

题意翻译

Feyn 喜欢玩括号序列。今天他想执行如下步骤 $n$ 次来构建一个括号序列： + 等概率随机选择一个空位（若当前有 $k$ 个字符，则有 $k+1$ 个空位）。 + 以 $p$ 的概率插入字符串 `()` 或以 $1-p$ 的概率插入字符串 `)(`，操作后字符串长度增加 $2$。 给定 $n,p$，求出 Feyn 得到一个合法括号序列的概率，对 $998244353$ 取模。 **注意**：读入的是 $q$，而 $p=q\times 10^{-4}$ （translated by 342873）

题目描述

Vika likes playing with bracket sequences. Today she wants to create a new bracket sequence using the following algorithm. Initially, Vika's sequence is an empty string, and then she will repeat the following actions $ n $ times: - Choose a place in the current bracket sequence to insert new brackets uniformly at random. If the length of the current sequence is $ k $ , then there are $ k+1 $ such places: before the first bracket, between the first and the second brackets, $ \ldots $ , after the $ k $ -th bracket. In particular, there is one such place in an empty bracket sequence. - Choose string "()" with probability $ p $ or string ")(" with probability $ 1 - p $ and insert it into the chosen place. The length of the bracket sequence will increase by $ 2 $ . A bracket sequence is called regular if it is possible to obtain a correct arithmetic expression by inserting characters '+' and '1' into it. For example, sequences "(())()", "()", and "(()(()))" are regular, while ")(", "(()", and "(()))(" are not. Vika wants to know the probability that her bracket sequence will be a regular one at the end. Help her and find this probability modulo $ 998\,244\,353 $ (see Output section).

输入输出格式

输入格式

The only line contains two integers $ n $ and $ q $ ( $ 1 \le n \le 500 $ ; $ 0 \le q \le 10^4 $ ). Here $ n $ is equal to the number of bracket insertion operations, and the probability that Vika chooses string "()" on every step of the algorithm is equal to $ p = q \cdot 10^{-4} $ .

输出格式

Print the probability that Vika's final bracket sequence will be regular, modulo $ 998\,244\,353 $ . Formally, let $ M = 998\,244\,353 $ . It can be shown that the answer can be expressed as an irreducible fraction $ \frac{p}{q} $ , where $ p $ and $ q $ are integers and $ q \not \equiv 0 \pmod{M} $ . Output the integer equal to $ p \cdot q^{-1} \bmod M $ . In other words, output such an integer $ x $ that $ 0 \le x < M $ and $ x \cdot q \equiv p \pmod{M} $ .

输入输出样例

输入样例 #1

1 7500

输出样例 #1

249561089

输入样例 #2

2 6000

输出样例 #2

519087064

输入样例 #3

5 4000

输出样例 #3

119387743

说明

In the first example, Vika will get a regular bracket sequence () with probability $ p = \frac{3}{4} $ , and she will get an irregular bracket sequence )( with probability $ 1 - p = \frac{1}{4} $ . The sought probability is $ \frac{3}{4} $ , and $ 249\,561\,089 \cdot 4 \equiv 3 \pmod{998\,244\,353} $ . In the second example, the sought probability is $ \frac{11}{25} $ .

题意翻译

Feyn 喜欢玩括号序列。今天他想执行如下步骤 $n$ 次来构建一个括号序列： + 等概率随机选择一个空位（若当前有 $k$ 个字符，则有 $k+1$ 个空位）。 + 以 $p$ 的概率插入字符串 `()` 或以 $1-p$ 的概率插入字符串 `)(`，操作后字符串长度增加 $2$。 给定 $n,p$，求出 Feyn 得到一个合法括号序列的概率，对 $998244353$ 取模。 **注意**：读入的是 $q$，而 $p=q\times 10^{-4}$ （translated by 342873）