Add "reference" for EN version. Bug fixes. (#1326)

en/docs/chapter_greedy/fractional_knapsack_problem.md

@@ -39,9 +39,9 @@ Since an `Item` object list is initialized, **the space complexity is $O(n)$**.
 
 ### Correctness proof
 
-Using proof by contradiction. Suppose item $x$ has the highest unit value, and some algorithm yields a maximum value `res`, but the solution does not include item $x`.
+Using proof by contradiction. Suppose item $x$ has the highest unit value, and some algorithm yields a maximum value `res`, but the solution does not include item $x$.
 
-Now remove a unit weight of any item from the knapsack and replace it with a unit weight of item $x$. Since the unit value of item $x$ is the highest, the total value after replacement will definitely be greater than `res`. **This contradicts the assumption that `res` is the optimal solution, proving that the optimal solution must include item $x**.
+Now remove a unit weight of any item from the knapsack and replace it with a unit weight of item $x$. Since the unit value of item $x$ is the highest, the total value after replacement will definitely be greater than `res`. **This contradicts the assumption that `res` is the optimal solution, proving that the optimal solution must include item $x$**.
 
 For other items in this solution, we can also construct the above contradiction. Overall, **items with greater unit value are always better choices**, proving that the greedy strategy is effective.
 

en/docs/chapter_greedy/max_product_cutting_problem.md

@@ -1,4 +1,4 @@
-# Maximum product after cutting problem
+# Maximum product cutting problem
 
 !!! question
 
@@ -40,7 +40,7 @@ As shown below, when $n \geq 4$, splitting out a $2$ increases the product, **wh
 
 Next, consider which factor is optimal. Among the factors $1$, $2$, and $3$, clearly $1$ is the worst, as $1 \times (n-1) < n$ always holds, meaning splitting out $1$ actually decreases the product.
 
-As shown below, when $n = 6$, $3 \times 3 > 2 \times 2 \times 2$. **This means splitting out $3$ is better than splitting out $2**.
+As shown below, when $n = 6$, $3 \times 3 > 2 \times 2 \times 2$. **This means splitting out $3$ is better than splitting out $2$**.
 
 **Greedy strategy two**: In the splitting scheme, there should be at most two $2$s. Because three $2$s can always be replaced by two $3$s to obtain a higher product.