結果

問題 No.3147 Parentheses Modification and Rotation (RM Ver.)
ユーザー 👑 binap
提出日時 2025-05-16 22:41:02
言語 C++17
(gcc 13.3.0 + boost 1.87.0)
結果
AC  
実行時間 31 ms / 2,000 ms
コード長 2,880 bytes
コンパイル時間 3,620 ms
コンパイル使用メモリ 255,884 KB
実行使用メモリ 9,000 KB
最終ジャッジ日時 2025-05-16 22:41:08
合計ジャッジ時間 5,091 ms
ジャッジサーバーID
（参考情報） 		judge4 / judge5
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 * 2
other AC * 33
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);}

using S = pair<int, int>;
S op(S a, S b){
	int d = min(a.second, b.first);
	return make_pair(a.first + (b.first - d), (a.second - d) + b.second);
}
S e(){
	return make_pair(0, 0);
}

int main(){
	int n;
	cin >> n;
	
	if(n % 2){
		cout << "-1\n";
		return 0;
	}
	
	string s;
	cin >> s;
	long long r, m;
	cin >> r >> m;
	
	segtree<S, op ,e> seg(2 * n);
	
	rep(i, n){
		if(s[i] == '('){
			seg.set(i, S(0, 1));
			seg.set(n + i, S(0, 1));
		}else{
			seg.set(i, S(1, 0));
			seg.set(n + i, S(1, 0));
		}
	}
	
	long long ans = 1001001001001001;
	
	rep(i, n){
		auto [left, right] = seg.prod(n - i, 2 * n - i);
		long long tmp = i * r + ((left + 1) / 2 + (right + 1) / 2) * m;
		chmin(ans, tmp);
//		cout << tmp << "\n";
	}
	
	cout << ans << "\n";
	
	return 0;
}
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