Submission #395973

#	Time	Username	Problem	Language	Result	Execution time	Memory
395973		amunduzbaev	Split the sequence (APIO14_sequence)	C++17	100 / 100	1970 ms	83804 KiB

sequence

/* made by amunduzbaev */
 
#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;
 
#define ff first
#define ss second
#define pb push_back
#define mp make_pair
#define ub upper_bound
#define lb lower_bound
#define sz(x) (int)x.size()
#define all(x) x.begin(), x.end()
#define rall(x) x.rbegin(),x.rend()
#define NeedForSpeed ios::sync_with_stdio(0);cin.tie(0);cout.tie(0);
#define vv vector
#define tut(x) array<int, x> 
//#define int long long
typedef long long ll;
typedef long double ld;
typedef pair<int, int> pii; 
typedef vector<int> vii;
typedef vector<pii> vpii;
 
template<class T> bool umin(T& a, const T& b) { return a > b? a = b, true:false; }
template<class T> bool umax(T& a, const T& b) { return a < b? a = b, true:false; }
//void usaco(string s) { freopen((s+".in").c_str(),"r",stdin);  
	//freopen((s+".out").c_str(),"w",stdout); }
//template<class T> tree<T, 
//less<T>, null_type, rb_tree_tag, 
//tree_order_statistics_node_update> ordered_set;
 
const int N = 1e5+5;
const int mod = 1e9+7;
const ll inf = 1e18;
const ld Pi = acos(-1);
const int B = 205;
 
#define MULTI 0
int n, m, K, s, ans, q, res, a[N];
int par[N][B], k, tt[N];
vv<ll> mx[2];
ll pref[N];
 
/*
 
7 3
4 1 3 4 0 2 3
 
*/
 
void dnc(int l = k, int r = n-1, int lx = k, int rx = n-1){
	int m = (l + r)>>1;
	mx[1][m] = -1;
	for(int i=lx;i<=min(rx, m);i++){
		if(umax(mx[1][m], mx[0][i-1] + (ll)(pref[m] - pref[i-1]) * (ll)pref[i-1])) par[m][k] = i;
	} if(l<=m-1) dnc(l, m-1, lx, par[m][k]);
	if(r>=m+1) dnc(m+1, r, par[m][k], rx);
}
 
void solve(int t_case) {
	NeedForSpeed
	cin>>n>>K;
	mx[1].resize(n), mx[0].resize(n);
	for(int i=0;i<n;i++) { cin>>a[i], pref[i] = a[i];
		if(i) pref[i] += pref[i-1]; }
	
	for(k=1;k<=K;k++) { dnc();
		swap(mx[1], mx[0]);
	} for(int k=K, in=n-1;k>0;k--) tt[k] = par[in][k], in = par[in][k] - 1;
	
	cout<<mx[0][n-1]<<"\n";
	for(int i=1;i<=K;i++) cout<<tt[i]<<" ";
	cout<<"\n";
}
 
signed main(){ 
	NeedForSpeed
	if(!MULTI) {
		solve(1);
	} else {
		int t; cin>>t;
		for(int t_case = 1; t_case <= t; t_case++) solve(t_case);
	} return 0;
}

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