update max_n

solutions/silver/cf-1077D.mdx

@@ -24,12 +24,12 @@ Once it is calculated, construct the resulting array by including elements in pr
 #include <iostream>
 #include <vector>
 
-const int MAX_N = 200001;
+const int MAX_N = 200000;
 
 int main() {
 	int n, k;
 	std::cin >> n >> k;
-	std::vector<int> freq(MAX_N);
+	std::vector<int> freq(MAX_N + 1);
 	std::vector<int> res;
 	for (int i = 0; i < n; i++) {
 		int x;
@@ -39,7 +39,7 @@ int main() {
 
 	std::function<bool(int)> can = [&](int x) {
 		int ele_num = 0;
-		for (int i = 1; i < MAX_N; i++) { ele_num += freq[i] / x; }
+		for (int i = 1; i <= MAX_N; i++) { ele_num += freq[i] / x; }
 		return (ele_num >= k);
 	};
 
@@ -53,7 +53,7 @@ int main() {
 		}
 	}
 
-	for (int i = 1; i < MAX_N; i++) {
+	for (int i = 1; i <= MAX_N; i++) {
 		for (int j = 0; j < freq[i] / lo; j++) { res.push_back(i); }
 	}
 
@@ -65,13 +65,13 @@ int main() {
 <PySection>
 
 ```py
-MAX_N = 200001
-freq = [0] * (MAX_N)
+MAX_N = 200000
+freq = [0] * (MAX_N + 1)
 
 
 def can(x: int) -> bool:
 	ele_num = 0
-	for i in range(1, MAX_N):
+	for i in range(1, MAX_N + 1):
 		ele_num += freq[i] // x
 
 	return ele_num >= k
@@ -94,7 +94,7 @@ while lo < hi:
 	else:
 		hi = mid - 1
 
-for i in range(1, MAX_N):
+for i in range(1, MAX_N + 1):
 	for j in range(0, freq[i] // lo):
 		res.append(i)