結果

問題 No.3129 Multiple of Twin Subarray
ユーザー 👑 binap
提出日時 2025-04-25 22:01:38
言語 C++17
(gcc 13.3.0 + boost 1.87.0)
結果
AC  
実行時間 84 ms / 2,000 ms
コード長 3,002 bytes
コンパイル時間 4,395 ms
コンパイル使用メモリ 253,208 KB
実行使用メモリ 17,436 KB
最終ジャッジ日時 2025-04-25 22:01:56
合計ジャッジ時間 7,897 ms
ジャッジサーバーID
(参考情報)
judge1 / 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 * 46
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);}

const long long INF = 100100101;

int main(){
	int n;
	cin >> n;
	vector<long long> a(n);
	cin >> a;
	
	auto f = [&](vector<long long> a){
		vector<long long> s(n + 1);
		rep(i, n) s[i + 1] = s[i] + a[i];
		long long MIN = 0, MAX = 0;
		vector<long long> p(n + 1);
		vector<long long> q(n + 1);
		p[0] = -INF;
		q[0] = INF;
		for(int x = 1; x <= n; x++){
			p[x] = max(s[x] - MIN, p[x - 1]);
			q[x] = min(s[x] - MAX, q[x - 1]);
			chmin(MIN, s[x]);
			chmax(MAX, s[x]);
		}
		return make_pair(p, q);
	};
	
	auto [p1, q1] = f(a);
	reverse(a.begin(), a.end());
	auto [p2, q2] = f(a);
	
	/*
	cout << p1;
	cout << q1;
	cout << p2;
	cout << q2;
	*/
	
	long long ans = -INF * INF;
	for(int x = 1; x <= n - 1; x++){
		chmax(ans, p1[x] * p2[n - x]);
		chmax(ans, p1[x] * q2[n - x]);
		chmax(ans, q1[x] * p2[n - x]);
		chmax(ans, q1[x] * q2[n - x]);
	}
	cout << ans << "\n";
	return 0;
}
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