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2 changes: 1 addition & 1 deletion C++/chapDynamicProgramming.tex
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Expand Up @@ -90,7 +90,7 @@ \subsubsection{分析}
\item 思路2：直接在i到j之间暴力枚举，复杂度是$O(n^3)$
\item 思路3：处理后枚举，连续子序列的和等于两个前缀和之差，复杂度$O(n^2)$。
\item 思路4：分治法，把序列分为两段，分别求最大连续子序列和，然后归并，复杂度$O(n\log n)$
\item 思路5：把思路2$O(n^2)$的代码稍作处理，得到$O(n)$的算法
\item 思路5：把思路3，$O(n^2)$的代码稍作处理，得到$O(n)$的算法
\item 思路6：当成M=1的最大M子段和
\myenddot

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