Submission #265388

#	Time	Username	Problem	Language	Result	Execution time	Memory
265388		amoo_safar	Split the sequence (APIO14_sequence)	C++17	100 / 100	1472 ms	85112 KiB

sequence

#include <bits/stdc++.h>
 
#define pb push_back
#define F first
#define S second
#define all(x) x.begin(), x.end()
#define debug(x) cerr << #x << " : " << x << '\n'
 
using namespace std;
 
typedef long long ll;
typedef long double ld;
typedef string str;
typedef pair<ll, ll> pll;
 
const ll Mod = 1e9 + 7;
//const ll Inf = 2242545357980376863LL;
const ll Log = 20;
//const ll N = 1ll << Log;
const int Maxn = 1e5 + 10;
//const int Base = 101;
 
 

typedef pair<ll, ll> Line;
typedef pair<pair<ll, Line>, int> Ty;

vector<Ty> L;

ll INF = 4e18;

inline ll Intersection(Line X, Line Y){
	if (X.first == Y.first && X.second >= Y.second)
		return (-INF);
	if (X.first == Y.first)
		return (INF);
	return ((Y.second - X.second) / (X.first - Y.first)) + ((Y.second - X.second) % (X.first - Y.first) > 0);
}
inline void Add(Line X, ll idx){
	while(!L.empty() && Intersection(L.back().F.S, X) <= L.back().F.F)
		L.pop_back();
	L.pb({{(L.empty() ? -INF : Intersection(L.back().F.S, X)), X}, idx});
}
inline pll GetMin(ll X){
	auto it = --upper_bound(all(L), Ty({X, {INF, INF}}, INF));// it --;
	return pll((X * it->F.S.F + it->F.S.S), it -> S);
}
 
 
int a[Maxn];//, mk[Maxn];
ll ps[Maxn];//, sf[Maxn];
 
ll dp[2][Maxn];
int par[203][Maxn];
vector<ll> A;
 
int main(){
	ios::sync_with_stdio(0); cin.tie(0); cout.tie(0);
	L.reserve(Maxn);
	//debug((-3) / 2);
	ll n, k;
	cin >> n >> k;
	for(int i = 0; i < n; i++) cin >> a[i];
	ps[0] = 0;
	for(int i = 1; i <= n; i++) ps[i] = ps[i - 1] + a[i - 1];
	
	for(int i = 0; i <= n; i++) dp[0][i] = ps[i] * ps[i];
	pll res;
	int jj;
	for(int j = 1; j <= k; j++){
		L.clear();
		jj = j & 1;

		dp[jj][0] = 0;
		for(int i = 1; i <= n; i++){
			Add( pll(-2LL * ps[i - 1], dp[jj ^ 1][i - 1] + ps[i - 1]*ps[i - 1] ), i - 1);
			res = GetMin(ps[i]);
			dp[jj][i] = res.F + ps[i]*ps[i];
			par[j][i] = res.S;
		}
	}
	cout << ((ps[n] * ps[n]) - dp[k&1][n]) / 2LL << '\n';
	ll nw = n;
	for(int i = 0; i < k; i++){
		nw = par[k - i][nw];
		A.pb(nw);
	}
	reverse(all(A));
	for(auto x : A) cout << x << ' ';
	cout << '\n';
	return 0;
}
/*
7 3
4 1 3 4 0 2 3

*/

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