제출 #723524

#	제출 시각	아이디	문제	언어	결과	실행 시간	메모리
723524		junseo	수열 (APIO14_sequence)	C++17	62 / 100	683 ms	81612 KiB

sequence

#include <bits/stdc++.h>
#define fi first
#define se second
#define eb emplace_back
#define ep emplace
#define all(v) (v).begin(), (v).end()
#define sz(v) (int)v.size()
#define rmin(r, x) r = min(r, x)
#define rmax(r, x) r = max(r, x)
#define ends ' '
#define endl '\n'
#define rep(i, s, e) for(int i = (s); i <= (e); ++i)
#define rep2(i, e, s) for(int i = (e); i >= (s); --i)
#define print(v) copy(all(v), ostream_iterator<int>(cout, " ")), cout << endl
#define fastio ios_base::sync_with_stdio(0), cin.tie(0)
using namespace std;
typedef long long ll;
typedef pair<int, int> pii;
typedef pair<ll, ll> pll;
typedef long double ld;
typedef vector<int> vi;

// #define LOCAL

const ll inf = 1e18;
const int maxn = 1e5 + 10;
const int maxk = 2e2 + 10;

int N, K;
ll S[maxn];
ll D[2][maxn];
int P[maxk][maxn];

struct Line {
	ll a, b;
	int prv;
	Line(ll a=0, ll b=-inf, int prv=-1) : a(a), b(b), prv(prv) {}
	ll calc(ll x) { return a * x + b; }
};

struct CHT {
	deque<Line> f;
	void update(Line g) {
		// while(not f.empty() and f.back().a == g.a) {
		// 	if(f.back().b > g.b)	return;
		// 	f.pop_back();
		// }
		if(not f.empty() and f.back().a == g.a and f.back().b >= g.b)	return;
		while(sz(f) >= 2) {
			Line& t2 = f[sz(f)-2];
			Line& t1 = f[sz(f)-1];
			// point(2, 0) < point(1, 0): break
			if((__int128)(t1.b-t2.b)*(t2.a-g.a) < (__int128)(g.b-t2.b)*(t2.a-g.a))	break;
			f.pop_back();
		}
		f.eb(g);
	}
	pll query(ll x) {
		if(f.empty())	return pll(-inf, -1);
		while(sz(f) >= 2 and f[0].calc(x) <= f[1].calc(x))	f.pop_front();
		return pll(f.front().calc(x), f.front().prv);
	}
} cht;

int main() {
	#ifdef LOCAL
		freopen("input.txt", "r", stdin);
		freopen("output.txt", "w", stdout);
	#endif
	fastio;

	cin >> N >> K;
	rep(i, 1, N) {
		int x;	cin >> x;
		S[i] = S[i-1] + x;
	}
	rep(i, 1, N)	P[0][i] = i;
	rep(k, 1, K) {
		int t = k & 1, _t = !t;
		CHT cht;
		rep(i, 1, N) {
			auto [d, p] = cht.query(S[i]);
			D[t][i] = d;
			P[k][i] = p;
			cht.update(Line(S[i], D[_t][i] - S[i]*S[i], i));
		}
	}
	vector<int> ans;
	int idx = N;
	rep2(k, K, 1) {
		idx = P[k][idx];
		ans.eb(idx);
	}
	reverse(all(ans));

	cout << D[K&1][N] << endl;
	for(auto& i : ans)	cout << i << ends;
	return 0;
}

• Subtask 1: 11 / 11
• Subtask 2: 11 / 11
• Subtask 3: 11 / 11
• Subtask 4: 0 / 17
• Subtask 5: 0 / 21
• Subtask 6: 29 / 29