\documentclass[11pt]{article}
\usepackage{CJK}
\textwidth 6.3in
\evensidemargin 0in
\oddsidemargin 0in
\begin{document}

\begin{CJK*}{Bg5}{ming}

\begin{enumerate}
\item 起始時\ $n$ 個\ processor $P_k$ 上有
$F = \left[
\begin{array}{c}
F_0\\
F_1\\
\end{array}
\right] =
\left[
\begin{array}{c}
1\\
1\\
\end{array}
\right]
,
M_k =
\left[
\begin{array}{cc}
0 & 1\\
1 & 1\\
\end{array}
\right]
\forall k \in \{1 \dots n\}$
\item 令\ $i = 0$
\item 若\ $k > 2^i$ 則\ $P_k$ 向\ $P_{k - 2^i}$ 取得\ $M_{k - 2^i}, \forall k \in \{1, \dots, n\},$ 若\ $2^i \ge n$ 則到步驟\ 6
\item $P_k$ 計算\ $M_{k - 2^i} * M_k$ 並將原來的\ $M_k$ 替換為計算出來的新值
\item $i = i + 1,$ 回到步驟\ 3
\item $P_k$ 計算\ $M_k * F$ 可得
$\left[
\begin{array}{c}
F_k\\
F_{k+1}\\
\end{array}
\right]
\forall k \in \{1, \dots, n\}$
\item 結束，processor $P_k$ 上有\ $F_k$ 之值\ $\forall k \in \{1, \dots, n\}$

\end{enumerate}

\end{CJK*}
\end{document}