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🔍 Filter Rules:
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html log:
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STDOUT: 
STDERR: WARNING: pandoc-crossref was compiled with pandoc 3.6.2 but is being run through 3.6.4. This is not supported. Strange things may (and likely will) happen silently.

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  \hypersetup{
    pdftitle={「杂题记录」括号序列（格路计数）},
    pdfauthor={Jiayi Su (ShuYuMo)},
    hidelinks,
    pdfcreator={LaTeX via pandoc}}
  
  \title{「杂题记录」括号序列（格路计数）}
  \author{Jiayi Su (ShuYuMo)}
  \date{2021-01-19 15:35:06}
  
  \begin{document}
  \maketitle
  
  \setstretch{1.3}
  给出两个正整数 \(n, k\) 。
  
  求出有多少个长度为 \(n\) 的括号序列，满足最长合法括号子序列长度恰好为
  \(2k\) 。
  
  \subsection{分析}\label{ux5206ux6790}
  
  根据\href{./「琐记」卡特兰数}{卡特兰数的转化}同样可以对这个问题进行转化。
  
  可以发现如果前缀和为 \(S_i\) ，那么最长的合法括号子序列长度为
  \(n-S_n + 2\min\{S\}\)，注意 \(min\{S\}\)通常为负值。
  
  \begin{figure}
  \centering
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  \caption{example.png}
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  其中红色的线段为无效符号，E 为最低点。
  
  设 \(t=\min\{S\}\)，易知 \(S_n = n+2t+2k\)。
  
  可以考虑枚举 \(t\) ，问题就变成了格路计数，需要满足一定到达过 \(y=t\)
  这个直线，且未曾穿过。
  
  未曾穿过的限制可以考虑用\href{./「琐记」卡特兰数}{卡特兰数的推导}相同的思想，即翻折引理。
  
  到达过这条线的要求可以考虑用 \(\min\{S\} \ge t\) 的答案减去
  \(\min\{S\} > t\) 的答案得到。
  
  考虑不经过 \(y=t-1\) 时的答案：（根据翻折引理） \[
  \dbinom{n}{(n-S_n)/2}-\dbinom{n}{\left[n - (2(t-1)-S_n)\right]/2}=\dbinom{n}{k-t}-\dbinom{n}{k-1}
  \] 同理，不经过 \(y=t\) 的答案为： \[
  \dbinom{n}{k-t}-\dbinom{n}{k}
  \]
  
  做差发现是： \[
  \dbinom{n}{k-1}-\dbinom{n}{k}
  \] 有不等式 \(S_n \ge \min\{S\}\)，易知：\(t \ge 2k-n\).即 \(k\) 有
  \(n-2k+1\) 种取值。
  
  答案为： \((n+1-2k)\left(\dbinom{n}{k}-\dbinom{n}{k-1}\right)\)
  
  这应该就能做到线性了，如果有个小巧的质数就能做到更快了。
  
  \end{document}
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