Submission #544247

#	Time	Username	Problem	Language	Result	Execution time	Memory
544247		SavicS	Split the sequence (APIO14_sequence)	C++17	0 / 100	60 ms	131072 KiB

sequence

#include <ext/pb_ds/assoc_container.hpp>
#include <ext/pb_ds/tree_policy.hpp>
#include <bits/stdc++.h>
 
using namespace std;
using namespace __gnu_pbds;
 
typedef long long ll;
typedef long double ldb;
 
typedef pair<int,int> pii;
typedef pair<ll,ll> pll;
typedef pair<ldb,ldb> pdd;
 
#define ff(i,a,b) for(int i = a; i <= b; i++)
#define fb(i,b,a) for(int i = b; i >= a; i--)
#define trav(a,x) for (auto& a : x)
 
#define sz(a) (int)(a).size()
#define pb push_back
#define fi first
#define se second
#define lb lower_bound
#define ub upper_bound
#define all(a) a.begin(), a.end()
#define rall(a) a.rbegin(), a.rend()
 
mt19937 rng(chrono::steady_clock::now().time_since_epoch().count());

template<typename T>
using ordered_set = tree<T, null_type, less<T>, rb_tree_tag, tree_order_statistics_node_update>;

// os.order_of_key(k) the number of elements in the os less than k
// *os.find_by_order(k)  print the k-th smallest number in os(0-based)

const int mod = 1000000007;
const ll inf = 1e18 + 5;
const int mxN = 200005; 

int n, K;
int niz[mxN];

int pref[mxN];

ll dp[mxN][201];

struct line{
    ll k, b;
};

struct CHT{
    vector<line> hull;

    ldb cross(line x, line y){
        return (ldb) (x.b - y.b) / (ldb)(y.k - x.k);
    }   

    void add(line x){
        while(sz(hull) >= 2 && cross(x, hull.back()) <= cross(hull.back(), hull[sz(hull) - 2])){
            hull.pop_back();
        }
        hull.pb(x);
    }

    ll get(ll x){

        auto f = [&](int id){
            return hull[id].k * x + hull[id].b;
        };

        int l = 0, r = sz(hull) - 1;
        while(l < r){
            int mid = (l + r) / 2;
            if(f(mid) > f(mid + 1))r = mid;
            else l = mid + 1;
        }

        return f(l);
    }

};

int main() {
    cin.tie(0)->sync_with_stdio(0);

    cin >> n >> K; K += 1;
    ff(i,1,n)cin >> niz[i];

    ff(i,1,n)pref[i] = pref[i - 1] + niz[i];

    ff(i,0,n)ff(j,0,K)dp[i][j] = -inf;

    dp[0][0] = 0;
    ff(j,1,K){

        CHT cht; 
        ff(i,1,n){

            if(dp[i - 1][j - 1] != -inf){
                cht.add({pref[i - 1], dp[i - 1][j - 1] - 1ll * pref[i - 1] * pref[n]});
            }

            if(i >= j)dp[i][j] = cht.get(pref[i]) + 1ll * pref[i] * pref[n] - 1ll * pref[i] * pref[i];


        }
    }

    int i = n, j = K; vector<int> res;
    while(1){

        fb(l,i,1){
            if(dp[l - 1][j - 1] + 1ll * (pref[i] - pref[l - 1]) * (pref[n] - pref[i]) == dp[i][j]){
                i = l - 1; j -= 1;
                break;
            }
        }    

        if(i == 0)break;
        res.pb(i);

    }

    cout << dp[n][K] << '\n';
    for(auto c : res)cout << c << " ";
    cout << '\n';

    return 0;
}
/*
 
7 3 
4 1 3 4 0 2 3
 
// probati bojenje sahovski
*/
 
 
 
 
 

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