Merge pull request #5061 from Sosuke23/a

Cutting Out Editorial

content/3_Silver/Conclusion.problems.json

@@ -59,8 +59,7 @@
       "isStarred": false,
       "tags": ["Binary Search"],
       "solutionMetadata": {
-        "kind": "autogen-label-from-site",
-        "site": "CF"
+        "kind": "internal"
       }
     },
     {

solutions/silver/cf-1077D.mdx

@@ -0,0 +1,101 @@
+---
+id: cf-1077D
+source: CF
+title: Cutting Out
+author: Rameez Parwez
+---
+
+[Official Editorial (C++)] (https://codeforces.com/blog/entry/63274)
+
+## Explanation
+
+Use binary search to determine the maximum number of times the subarray $t$ can be cut out from the original array.
+Once it is calculated, construct the resulting array by including elements in proportion to their contributions.
+
+## Implementation
+
+**Time Complexity:** $\mathcal{O} (MAX\_N \log(N))$
+
+<LanguageSection>
+<CPPSection>
+
+```cpp
+#include <iostream>
+#include <vector>
+
+const int MAX_N = 200000;
+
+int main() {
+	int n, k;
+	std::cin >> n >> k;
+	std::vector<int> freq(MAX_N + 1);
+	for (int i = 0; i < n; i++) {
+		int x;
+		std::cin >> x;
+		freq[x] += 1;
+	}
+
+	auto can = [&](int x) -> bool {
+		int ele_num = 0;
+		for (int i = 1; i <= MAX_N; i++) { ele_num += freq[i] / x; }
+		return ele_num >= k;
+	};
+
+	int lo = 0;
+	int hi = n;
+	while (lo < hi) {
+		int mid = (lo + hi + 1) / 2;
+		if (can(mid)) {
+			lo = mid;
+		} else {
+			hi = mid - 1;
+		}
+	}
+
+	std::vector<int> res;
+	for (int i = 1; i <= MAX_N; i++) {
+		for (int j = 0; j < freq[i] / lo; j++) { res.push_back(i); }
+	}
+
+	for (int i = 0; i < k; i++) { std::cout << res[i] << " \n"[i == k - 1]; }
+}
+```
+
+</CPPSection>
+<PySection>
+
+```py
+MAX_N = 200000
+
+freq = [0] * (MAX_N + 1)
+n, k = map(int, input().split())
+for x in input().split():
+	freq[int(x)] += 1
+
+
+def can(x: int) -> bool:
+	ele_num = 0
+	for i in range(1, MAX_N + 1):
+		ele_num += freq[i] // x
+	return ele_num >= k
+
+
+lo = 0
+hi = n
+while lo < hi:
+	mid = (lo + hi + 1) // 2
+	if can(mid):
+		lo = mid
+	else:
+		hi = mid - 1
+
+res = []
+for i in range(1, MAX_N + 1):
+	for j in range(0, freq[i] // lo):
+		res.append(i)
+
+print(*res[:k])
+```
+
+</PySection>
+</LanguageSection>