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  \hypersetup{
    pdftitle={「杂题记录」「CTSC2017」吉夫特},
    pdfauthor={Jiayi Su (ShuYuMo)},
    hidelinks,
    pdfcreator={LaTeX via pandoc}}
  
  \title{「杂题记录」「CTSC2017」吉夫特}
  \author{Jiayi Su (ShuYuMo)}
  \date{2021-01-20 15:35:06}
  
  \begin{document}
  \maketitle
  
  \setstretch{1.3}
  给出一个长度为 \(n\) 的数列 \(A_i\) ,求有多少个长度 \(k\) 的子序列
  \(A'\) (\(k\ge 2\))满足: \[
  \prod_{i=1}^{k-1}\dbinom{A'_i}{A'_{i+1}} > 0 \pmod{2}
  \] \(n \le 211985, A_i \le 233333\)。原题保证 \(A_i\)
  互不相同,但是不重要。
  
  \subsection{分析}\label{ux5206ux6790}
  
  根据 Lucas 定理,就是求有多少 \(A\) 的子序列 \(A'\) 满足: \[
  \forall i \in [1, k-1],S(A_i) \subseteq S(A_{i+1})
  \] \(S(x)\) 表示二进制数 \(x\) 表示的集合。
  
  这东西可以直接 dp ,设 \(f(i)\) 表示以 \(i\) 结尾的合法子序列有多少: \[
  f(n) \sum_{i=1}^{n-1} f(i)[A_i \operatorname{and}A_n=A_n]
  \] 直接暴力枚举是 \(\mathcal{O}(n^2)\) 的。
  
  考虑类似于分块一样的优化,考虑将 \(A_i\) 拆开,设 \(A_i\) 二进制下的前 9
  位为 \(x\), 后 9 位为 \(y\)。 \(g(x, y)\) 表示前九位恰好为 \(x\)
  ,后九位是 \(y\) 的子集的 \(A_i\) 的对应 \(f(i)\) 之和。
  
  考虑维护这个东西,求出一个 \(g(i)\) 后枚举子集更新。
  
  考虑使用这个东西,在求一个 \(g(i)\) 时,枚举子集求出。
  
  成功均摊了复杂度。 总复杂度 \(\mathcal{O}(2^{9}n)\)
  
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  (.../mt-LatinModernRoman.cfg)
  (.../omllmm.fd)
  (.../umsa.fd)
  (.../mt-msa.cfg)
  (.../umsb.fd)
  (.../mt-msb.cfg)
  
  LaTeX Font Warning: Font shape `TU/ARPLUKaiCN(0)/b/n' undefined
  (Font)              using `TU/ARPLUKaiCN(0)/m/n' instead on input line 120.
  
  Missing character: There is no , (U+FF0C) in font LatinModernMath-Regular/OT:sc
  ript=math;language=dflt;!
  
  Package xeCJK Warning: Unknown CJK family `\CJKttdefault' is being ignored.
  (xeCJK)                
  (xeCJK)                Try to use `\setCJKmonofont[<...>]{<...>}' to define
  (xeCJK)                it.
  
  
  LaTeX Font Warning: Font shape `TU/ARPLUKaiCN(0)/m/it' undefined
  (Font)              using `TU/ARPLUKaiCN(0)/m/n' instead on input line 145.
  
  [1] [2] (.../input.aux)
  
  LaTeX Font Warning: Some font shapes were not available, defaults substituted.
  
   )
  Output written on .../input.pdf (2 pages).
  Transcript written on .../input.log.
[WARNING] Missing character: There is no , (U+FF0C) (U+FF0C) in font LatinModernMath-Regular/OT:sc