제출 #387129

#	제출 시각	아이디	문제	언어	결과	실행 시간	메모리
387129		idontreallyknow	수열 (APIO14_sequence)	C++17	49 / 100	116 ms	131076 KiB

sequence

#include <bits/stdc++.h>
#include <ext/pb_ds/assoc_container.hpp>
#include <ext/pb_ds/tree_policy.hpp>
using namespace std;
using namespace __gnu_pbds;
typedef long long ll;
template <class T> using ordered_set = tree<T, null_type, less<T>, rb_tree_tag, tree_order_statistics_node_update>;
mt19937 rng(chrono::steady_clock::now().time_since_epoch().count());
template<class A> ostream& operator<<(ostream &cout, vector<A> const &v) {cout << "["; for(int i = 0; i < v.size(); i++) {if (i) cout << ", "; cout << v[i];} return cout << "]";};
template<class A, class B> ostream& operator<<(ostream &cout, const pair<A,B> &x) {return cout << "(" <<x.first << ", " << x.second << ")";};
template <class T> void pv(T a, T b) {cerr << "["; for (T i = a; i != b; ++i) cerr << *i << " "; cerr << "]\n";}
void _print() {cerr << "]\n";}
template <class T, class... V> void _print(T t, V... v) {cerr << t; if (sizeof...(v)) cerr << ", "; _print(v...);}
#define debug(x...) cerr << "[" << #x << "] = ["; _print(x)
#define fi first
#define se second
#define SZ(x) (int)((x).size())
#define pii pair<int,int>
typedef long double ld;
const int mx = 1e5+5;
struct line {
	int idx;
	ll m, b;
	line(ll mm, ll bb, int ii) : idx(ii), m(mm), b(bb) {}
	ll eval(ll x) {return m*x+b;}
	ld isectx(line l) {return (ld)(l.b-b)/(m-l.m);}
};
ll a[mx], dp[mx][205];
int prv[mx][205];
deque<line> stuff[205];
int main() {
	ios::sync_with_stdio(0);
	cin.tie(0);
	int n,k;
	cin >> n >> k;
	for (int q = 1; q <= n; q++) cin >> a[q];
	stuff[0].push_back(line(0,0,0));
	ll cur = 0;
	for (int q = 1; q <= n; q++) {
		cur += a[q];
		for (int w = 1; w <= min(q,k+1); w++) {
			while (SZ(stuff[w-1]) >= 2 && stuff[w-1][0].eval(cur) >= stuff[w-1][1].eval(cur)) stuff[w-1].pop_front();
			dp[q][w] = stuff[w-1][0].eval(cur)+cur*cur;
			prv[q][w] = stuff[w-1][0].idx;
			line ins = line(cur*-2,dp[q][w]+cur*cur,q);
			while (SZ(stuff[w])) {
				if (stuff[w].back().m == ins.m) {
					if (ins.b < stuff[w].back().b) stuff[w].pop_back();
					else break;
				} else if (SZ(stuff[w]) >= 2) {
					if (ins.isectx(stuff[w].back()) <= ins.isectx(stuff[w][SZ(stuff[w])-2])) stuff[w].pop_back();
					else break;
				} else {
					break;
				}
			}
			if (SZ(stuff[w]) == 0 || stuff[w].back().m != ins.m) stuff[w].push_back(ins);
		}
	}
	cout << (cur*cur-dp[n][k+1])/2 << "\n";
	int ri = n, cuts = k+1;
	while (cuts > 1) {
		cout << prv[ri][cuts] << " ";
		ri = prv[ri][cuts];
		cuts--;
	}
	cout << "\n";
	return 0;
}

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