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Update H-laser-lab.tex #2

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6 changes: 3 additions & 3 deletions solutions/H-laser-lab.tex
Original file line number Diff line number Diff line change
Expand Up @@ -147,15 +147,15 @@ \section{H. 레이저 연구소}

\vspace{18pt}

$\displaystyle \sum_{i=1}^{n} f(i) = s_{f}(n), \displaystyle \sum_{i=1}^{n} g(i) = s_{g}(n), \displaystyle \sum_{i=1}^{n} f\ast g(i) = s_{f\astg}(n)$ 라고 하면
$\displaystyle \sum_{i=1}^{n} f(i) = s_{f}(n), \displaystyle \sum_{i=1}^{n} g(i) = s_{g}(n), \displaystyle \sum_{i=1}^{n} f\ast g(i) = s_{f\ast g}(n)$ 라고 하면

\vspace{18pt}

$s_{f}(n) = {{s_{f\astg}(n)-\displaystyle \sum_{d=2}^{n} s_{f}\left(\left\lfloor{n \over d}\right\rfloor\right)g(d)} \over {g(1)}}$가 됩니다.
$s_{f}(n) = {{s_{f\ast g}(n)-\displaystyle \sum_{d=2}^{n} s_{f}\left(\left\lfloor{n \over d}\right\rfloor\right)g(d)} \over {g(1)}}$가 됩니다.
\end{frame}

\begin{frame}{\textbf{H}. 레이저 연구소}
$s_{f\ast g}(n), s_{g}(n)$을 $O(1)$에 계산할 수 있다면 $s_{f}(n)$은 재귀적으로 $O(n^{3/4})$ 시간에 구할 수 있고, 구현에 따라 AC를 받을 수 있습니다.

$i \leq n^{2/3}$ 범위에서 $f(i)$를 전처리해두면 $O(n^{2/3})$ 시간에 문제를 풀 수 있고, 넉넉히 AC를 받을 수 있습니다.
\end{frame}
\end{frame}