fix: update

Author: yanglbme

solution/2000-2099/2052.Minimum Cost to Separate Sentence Into Rows/README.md

@@ -97,7 +97,7 @@ tags:
 函数 $\textit{dfs}(i)$ 的执行过程如下：
 
 -   如果从第 $i$ 个单词开始到最后一个单词的长度之和加上单词之间的空格数小于等于 $k$，那么这些单词可以放在最后一行，成本为 $0$。
--   否则，我们枚举下一个开始分隔的单词的位置 $j$，使得从第 $i$ 个单词到第 $j-1$ 个单词的长度之和加上单词之间的空格数小于等于 $k$。那么 $\textit{dfs}(j)$ 表示从第 $j$ 个单词开始分隔句子的最小成本，而 $\textit{dfs}(j) + (k - m)^2$ 表示将第 $i$ 个单词到第 $j-1$ 个单词放在一行的成本，其中 $m$ 表示从第 $i$ 个单词到第 $j-1$ 个单词的长度之和加上单词之间的空格数。我们枚举所有的 $j$，取最小值即可。
+-   否则，我们枚举下一个开始分隔的单词的位置 $j$，使得从第 $i$ 个单词到第 $j-1$ 个单词的长度之和加上单词之间的空格数小于等于 $k$。那么 $\textit{dfs}(j)$ 表示从第 $j$ 个单词开始分隔句子的最小成本，而 $(k - m)^2$ 表示将第 $i$ 个单词到第 $j-1$ 个单词放在一行的成本，其中 $m$ 表示从第 $i$ 个单词到第 $j-1$ 个单词的长度之和加上单词之间的空格数。我们枚举所有的 $j$，取最小值即可。
 
 答案即为 $\textit{dfs}(0)$。
 

solution/2000-2099/2052.Minimum Cost to Separate Sentence Into Rows/README_EN.md

@@ -96,7 +96,7 @@ Next, we design a function $\textit{dfs}(i)$, which represents the minimum cost
 The execution process of the function $\textit{dfs}(i)$ is as follows:
 
 -   If the sum of the lengths of the words from the $i$-th word to the last word plus the number of spaces between the words is less than or equal to $k$, then these words can be placed on the last line, and the cost is $0$.
--   Otherwise, we enumerate the position $j$ of the next word to start splitting, such that the sum of the lengths of the words from the $i$-th word to the $(j-1)$-th word plus the number of spaces between the words is less than or equal to $k$. Then $\textit{dfs}(j)$ represents the minimum cost of splitting the sentence starting from the $j$-th word, and $\textit{dfs}(j) + (k - m)^2$ represents the cost of placing the words from the $i$-th word to the $(j-1)$-th word on one line, where $m$ represents the sum of the lengths of the words from the $i$-th word to the $(j-1)$-th word plus the number of spaces between the words. We enumerate all $j$ and take the minimum value.
+-   Otherwise, we enumerate the position $j$ of the next word to start splitting, such that the sum of the lengths of the words from the $i$-th word to the $(j-1)$-th word plus the number of spaces between the words is less than or equal to $k$. Then $\textit{dfs}(j)$ represents the minimum cost of splitting the sentence starting from the $j$-th word, and $(k - m)^2$ represents the cost of placing the words from the $i$-th word to the $(j-1)$-th word on one line, where $m$ represents the sum of the lengths of the words from the $i$-th word to the $(j-1)$-th word plus the number of spaces between the words. We enumerate all $j$ and take the minimum value.
 
 The answer is $\textit{dfs}(0)$.