fix some issues

Author: SansPapyrus683

content/3_Silver/More_Prefix_Sums.mdx

@@ -20,16 +20,16 @@ frequency: 2
 ### Solution - Max Subarray Sum
 
 Consider the prefix sum array $p$. The subarray sum $a_i \dots a_{j-1}$, where
-$i\leq j$ is $p[j]-p[i]$. Thus, we are looking for the maximum possible value of
-$p[j]-p[i]$ over $0 \leq i \leq j \leq N$.
+$i < j$ is $p[j]-p[i]$. Thus, we are looking for the maximum possible value of
+$p[j]-p[i]$ over $0 \leq i < j \leq N$.
 
 For a fixed right bound $j$, the maximum subarray sum is
 
 $$
-p[j]-\min_{i\leq j}{p[i]}
+p[j]-\min_{i < j}{p[i]}
 $$
 
-Thus, we can keep a running minimum to store $\min\limits_{i\leq j}{p[i]}$ as we
+Thus, we can keep a running minimum to store $\min\limits_{i < j}{p[i]}$ as we
 iterate through $j$. This yields the maximum subarray sum for each possible
 right bound, and the maximum over all these values is our answer.
 
@@ -66,6 +66,7 @@ int main() {
 	cout << max_subarray_sum << endl;
 }
 ```
+
 </CPPSection>
 <JavaSection>
 

content/5_Plat/Binary_Jump.problems.json

@@ -238,7 +238,7 @@
     },
     {
       "uniqueId": "usaco-901",
-      "name": "Exercise",
+      "name": "Exercise Route",
       "url": "http://www.usaco.org/index.php?page=viewproblem2&cpid=901",
       "source": "Platinum",
       "difficulty": "Very Hard",