結果

問題 No.2930 Larger Mex
ユーザー 👑 binap
提出日時 2024-10-12 15:18:53
言語 C++17
(gcc 13.3.0 + boost 1.87.0)
結果
AC  
実行時間 130 ms / 2,000 ms
コード長 2,989 bytes
コンパイル時間 4,316 ms
コンパイル使用メモリ 252,436 KB
最終ジャッジ日時 2025-02-24 18:00:00
ジャッジサーバーID
（参考情報） 		judge3 / judge4
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 * 50
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 vector<int> vi;
typedef vector<long long> vl;
typedef vector<vector<int>> vvi;
typedef vector<vector<long long>> vvl;
typedef long double ld;
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, m;
	cin >> n >> m;
	vector<int> a(n);
	cin >> a;
	vector<vector<int>> memo(m);
	rep(i, n) if(a[i] < m) memo[a[i]].push_back(i);
	rep(i, m){
		if(int(memo[i].size()) == 0){
			rep(i, n) cout << "0\n";
			return 0;
		}
	}
	if(m == 0){
		rep(i, n) cout << n - i << "\n";
		return 0;
	}
	int r = 0;
	rep(i, m){
		chmax(r, memo[i][0]);
	}
	vector<int> ans(n + 1);
	rep(i, n){
		ans[r - i + 1]++;
		ans[n - i + 1]--;
		if(a[i] >= m) continue;
		auto it = upper_bound(memo[a[i]].begin(), memo[a[i]].end(), i);
		if(it == memo[a[i]].end()) r = n;
		else chmax(r, *it);
	}
	int sum = 0;
	rep(i, n + 1){
		ans[i] += sum;
		sum = ans[i];
	}
	for(int i = 1; i <= n; i++) cout << ans[i] << "\n";
	return 0;
}
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