fix: md

solution/0600-0699/0638.Shopping Offers/README.md

@@ -76,7 +76,7 @@ tags:
 
 我们注意到，题目中物品的种类 $n \leq 6$，而每种物品需要购买的数量不超过 $10$，我们可以用 $4$ 个二进制位来表示每种物品需要购买的数量，这样，我们只需要最多 $6 \times 4 = 24$ 个二进制位来表示整个购物清单。
 
-我们首先将购物清单 $\text{needs}$ 转换为一个整数 $\text{mask}$，其中第 $i$ 个物品需要购买的数量存储在 $\text{mask}$ 的第 $i \times 4$ 位到第 $(i + 1) \times 4 - 1$ 位。例如，当 $\text{needs} = [1, 2, 1]$ 时，有 $\text{mask} = 0b0001\_0010\_0001$。
+我们首先将购物清单 $\text{needs}$ 转换为一个整数 $\text{mask}$，其中第 $i$ 个物品需要购买的数量存储在 $\text{mask}$ 的第 $i \times 4$ 位到第 $(i + 1) \times 4 - 1$ 位。例如，当 $\text{needs} = [1, 2, 1]$ 时，有 $\text{mask} = 0b0001 0010 0001$。
 
 然后，我们设计一个函数 $\text{dfs}(cur)$，表示当前购物清单的状态为 $\text{cur}$ 时，我们需要花费的最少金额。那么答案即为 $\text{dfs}(\text{mask})$。
 

solution/0600-0699/0638.Shopping Offers/README_EN.md

@@ -74,7 +74,7 @@ You cannot add more items, though only $9 for 2A ,2B and 1C.
 
 We notice that the number of types of items $n \leq 6$ in the problem, and the quantity of each item needed does not exceed $10$. We can use $4$ binary bits to represent the quantity of each item needed. Thus, we only need at most $6 \times 4 = 24$ binary bits to represent the entire shopping list.
 
-First, we convert the shopping list $\text{needs}$ into an integer $\text{mask}$, where the quantity of the $i$-th item needed is stored in the $i \times 4$ to $(i + 1) \times 4 - 1$ bits of $\text{mask}$. For example, when $\text{needs} = [1, 2, 1]$, we have $\text{mask} = 0b0001\_0010\_0001$.
+First, we convert the shopping list $\text{needs}$ into an integer $\text{mask}$, where the quantity of the $i$-th item needed is stored in the $i \times 4$ to $(i + 1) \times 4 - 1$ bits of $\text{mask}$. For example, when $\text{needs} = [1, 2, 1]$, we have $\text{mask} = 0b0001 0010 0001$.
 
 Then, we design a function $\text{dfs}(cur)$, representing the minimum amount of money we need to spend when the current state of the shopping list is $\text{cur}$. Therefore, the answer is $\text{dfs}(\text{mask})$.