結果

問題 No.3050 Prefix Removal
コンテスト
ユーザー 👑 binap
提出日時 2025-03-07 22:27:48
言語 C++17
(gcc 13.3.0 + boost 1.87.0)
結果
AC  
実行時間 472 ms / 2,000 ms
コード長 3,922 bytes
コンパイル時間 4,422 ms
コンパイル使用メモリ 254,420 KB
実行使用メモリ 30,684 KB
最終ジャッジ日時 2025-03-07 22:28:11
合計ジャッジ時間 21,375 ms
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
other AC * 55
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);}

// less : #SMALLESTS <= Size
// greater : #LARGESTS <= Size
template<typename T = long long, typename U = less<long long>>
struct DoubleMultisetSize{
	static_assert(is_same<U, std::greater<T>>::value || is_same<U, std::less<T>>::value, "U must be either std::greater<T> or std::less<T>");
	int k;
	multiset<T, U> OK, NG;
	T sum_OK, sum_NG;
	DoubleMultisetSize(int k) : k(k), sum_OK(0), sum_NG(0) {};
	void balance(){
		while(int(OK.size()) > k){
			auto it = prev(OK.end());
    		sum_NG += *it;
    		sum_OK -= *it;
    		NG.insert(*it);
    		OK.erase(it);
    	}
		while(int(NG.size()) > 0 and int(OK.size()) < k){
			auto it = NG.begin();
    		sum_OK += *it;
    		sum_NG -= *it;
    		OK.insert(*it);
    		NG.erase(it);
    	}
	}
	void add(T x){
		OK.insert(x);
		sum_OK += x;
		balance();
	}
	void del(T x){
		if(OK.find(x) == OK.end()){
			sum_NG -= x;
			NG.erase(NG.find(x));
		}else{
			sum_OK -= x;
			OK.erase(OK.find(x));
			balance();
		}
	};
	void change_size(int k_new){
		k = k_new;
		balance();
	}
	pair<int, int> get_size(){return make_pair(OK.size(), NG.size());}
	pair<T, T> get_sum(){return make_pair(sum_OK, sum_NG);}
};

int main(){
	int n, k;
	cin >> n >> k;
	vector<long long> a(n);
	cin >> a;
	
	DoubleMultisetSize<long long, greater<long long>> dm(k - 1);
	long long offset = 0;
	long long ans = -4004004004004004004;
	rep(i, n){
		offset += a[i];
		if(i > 0) dm.add(-offset + a[i]);
		if(i >= k - 1){
			long long tmp = dm.get_sum().first;
//			for(long long x : dm.OK) cout << x << ' ';
//			cout << "\n";
			tmp += offset * k;
//			cout << tmp << "\n";
			chmax(ans, tmp);
		}
	}
	cout << ans << "\n";
	
	return 0;
}
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