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pdftitle={「杂题记录」「CTSC2017」吉夫特},
pdfauthor={Jiayi Su (ShuYuMo)},
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\title{「杂题记录」「CTSC2017」吉夫特}
\author{Jiayi Su (ShuYuMo)}
\date{2021-01-20 15:35:06}
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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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