Update cf-100886G.mdx

Author: SansPapyrus683

solutions/gold/cf-100886G.mdx

@@ -6,42 +6,50 @@ author: Daniel Zhu
 ---
 
 <Spoiler title="Hint">
+
 If $l = 1$, what will the best solution always be?
+
 </Spoiler>
+
 ## Explanation
+
 ### $l = 1$
+
 Let's start with an example: $l = 1$ and $r = 875$. Observe that we really need to consider three numbers as viable answers: $875$, $799$, and $869$. Try to think about why, and how this generalizes!
+
 ### $l \neq 1$
+
 We can apply a very similar approach when $l \neq 1$. In fact, it's almost identical, with one minor difference. Consider the case where $l = 855$ and $r = 875$, just as before. The only difference is we can no longer consider $799$ as a viable answer, as it is less than $l$. Again, try to think about how to generalize this!
 
 ## Implementation
 
-**Time Complexity:** $\mathcal{O}(D^2)$
-
-_Note:_ $D$ is the max number of digits (19).
+**Time Complexity:** $\mathcal{O}(D^2)$, where $D$ is the maximum number of digits (19).
 
 <LanguageSection>
-
 <CPPSection>
 
 ```cpp
 #include <bits/stdc++.h>
 using ll = long long;
 using namespace std;
-/** @return product of the given string's digits (-1 if empty) */
+
+/** @return the product of the given string's digits (-1 if empty) */
 ll prod(string s) {
 	if (!s.length()) return -1;
 	ll res = 1;
 	for (char c : s) { res *= c - '0'; }
 	return res;
 }
+
 int main() {
 	string l, r;
 	cin >> l >> r;
+
 	// pad beginning of l with zeros
 	// l = 12, r = 132 -> l = 012, r = 132
 	while (l.length() < r.length()) { l.insert(l.begin(), '0'); }
 	string ans = "";
+
 	// i -> length of LCP (longest common prefix) of str and r
 	bool eq = true;  // whether l[0, i) = r[0, i)
 	for (int i = 0; i <= r.length(); i++) {
@@ -58,12 +66,12 @@ int main() {
 		cur += string(r.length() - cur.length(), '9');
 		if (prod(cur) > prod(ans)) { ans = cur; }
 	}
+
 	// remove any leading zeros from ans
 	while (ans[0] == '0') { ans.erase(ans.begin()); }
 	cout << ans << endl;
 }
 ```
 
 </CPPSection>
-
 </LanguageSection>