結果

問題 No.3269 Leq-K Partition
コンテスト
ユーザー 👑 binap
提出日時 2025-09-12 23:28:09
言語 C++17
(gcc 13.3.0 + boost 1.87.0)
結果
TLE  
実行時間 -
コード長 2,949 bytes
コンパイル時間 4,020 ms
コンパイル使用メモリ 257,948 KB
実行使用メモリ 14,112 KB
最終ジャッジ日時 2025-09-12 23:45:41
合計ジャッジ時間 12,036 ms
ジャッジサーバーID
（参考情報） 		judge10 / judge3
このコードへのチャレンジ
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ファイルパターン 結果
sample AC * 3
other AC * 5 TLE * 1 -- * 21
権限があれば一括ダウンロードができます

ソースコード

diff # 

#include<bits/stdc++.h>
#include<atcoder/all>
#define rep(i,n) for(int i=0;i<n;i++)
using namespace std;
using namespace atcoder;
typedef long long ll;
typedef pair<int, int> P;

template <int m> ostream& operator<<(ostream& os, const static_modint<m>& a) {os << a.val(); return os;}
template <int m> ostream& operator<<(ostream& os, const dynamic_modint<m>& a) {os << a.val(); return os;}
template <int m> istream& operator>>(istream& is, static_modint<m>& a) {long long x; is >> x; a = x; return is;}
template <int m> istream& operator>>(istream& is, dynamic_modint<m>& a) {long long x; is >> x; a = x; return is;}
template<typename T> istream& operator>>(istream& is, vector<T>& v){int n = v.size(); assert(n > 0); rep(i, n) is >> v[i]; return is;}
template<typename U, typename T> ostream& operator<<(ostream& os, const pair<U, T>& p){os << p.first << ' ' << p.second; return os;}
template<typename T> ostream& operator<<(ostream& os, const vector<T>& v){int n = v.size(); rep(i, n) os << v[i] << (i == n - 1 ? "\n" : " "); return os;}
template<typename T> ostream& operator<<(ostream& os, const vector<vector<T>>& v){int n = v.size(); rep(i, n) os << v[i] << (i == n - 1 ? "\n" : ""); return os;}
template<typename T> ostream& operator<<(ostream& os, const set<T>& se){for(T x : se) os << x << " "; os << "\n"; return os;}
template<typename T> ostream& operator<<(ostream& os, const unordered_set<T>& se){for(T x : se) os << x << " "; os << "\n"; return os;}
template<typename S, auto op, auto e> ostream& operator<<(ostream& os, const atcoder::segtree<S, op, e>& seg){int n = seg.max_right(0, [](S){return true;}); rep(i, n) os << seg.get(i) << (i == n - 1 ? "\n" : " "); return os;}
template<typename S, auto op, auto e, typename F, auto mapping, auto composition, auto id> ostream& operator<<(ostream& os, const atcoder::lazy_segtree<S, op, e, F, mapping, composition, id>& seg){int n = seg.max_right(0, [](S){return true;}); rep(i, n) os << seg.get(i) << (i == n - 1 ? "\n" : " "); return os;}

template<typename T> void chmin(T& a, T b){a = min(a, b);}
template<typename T> void chmax(T& a, T b){a = max(a, b);}

int main(){
	int n;
	cin >> n;
	vector<int> a(n);
	cin >> a;
	rep(i, n) a[i]--;
	const int M = min(n, int(sqrt(n)));
	vector<int> ans(n + 1, -1);
	auto solve = [&](int k){
		int res = 0;
		int cnt = 0;
		set<int> se;
		rep(i, n){
			se.insert(a[i]);
			if(se.size() > k){
				se.clear();
				res++;
				se.insert(a[i]);
			}
		}
		return res + 1;
	};
	
	for(int k = 1; k <= M; k++){
		auto res = solve(k);
		ans[k] = res;
	}
	int start = M;
	while(start <= n){
		int ok = start, ng = n + 1;
		int res_base = solve(ok);
		
		while(ng - ok > 1){
			int mid = (ok + ng) / 2;
			int res = solve(mid);
			if(res == res_base) ok = mid;
			else ng = mid;
		}
		if(ng <= n)	ans[ng] = solve(ng);
		start = ng;
	}
	for(int k = 1; k <= n; k++){
		if(k >= 2) if(ans[k] == -1) ans[k] = ans[k - 1];
		cout << ans[k] << "\n";
	}
	return 0;
}
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