Submission #983052

#	Time	Username	Problem	Language	Result	Execution time	Memory
983052		Hugo1729	Split the sequence (APIO14_sequence)	C++11	49 / 100	70 ms	131072 KiB

sequence

#include <bits/stdc++.h>
using namespace std;
typedef long long ll;

ll dp[100001][202]={0};
ll pref[100001]={0};
ll p[100001][202];

struct line{
    ll m,c;
 
    line(ll _m, ll _c){m=_m;c=_c;}
 
    ll intersect(line l){
        return m-l.m==0 ? LLONG_MAX/4 : (l.c-c)/(m-l.m);
    }
 
    ll value(ll x){
        return m*x+c;
    }
};
 
deque<pair<line,ll>> q;
deque<ll> pos;
 
void insert(ll m, ll c){
    line l(m,c);
    while(!q.empty()&&q.front().second>l.intersect(q.front().first)){
        // cout << "irem" << ' ' << q.front().first.m << ' ' << q.front().first.c << ' ' << q.front().second << '\n';
        q.pop_front();
        pos.pop_front();
    }
 
    if(q.empty()){
        q.push_front({l,0});
    }else{
        q.push_front({l,l.intersect(q.front().first)});
    }
    // cout << "iadd" << ' ' << q.front().first.m << ' ' << q.front().first.c << ' ' << q.front().second << '\n';
}
 
ll query(ll x){
    while(q.size()>1&&q[q.size()-2].second<x){
        // cout << "qrem" << ' ' << q.back().first.m << ' ' << q.back().first.c << ' ' << q.back().second << '\n';
        q.pop_back();
        pos.pop_back();
    }
 
    // cout << "query" << ' ' << q.back().first.m << ' ' << q.back().first.c << ' ' << q.back().second <<  ' ' << q.back().first.value(x) <<'\n';
 
    return q.back().first.value(x);
}

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

    int n, k; cin >> n >> k;
    vector<ll> a(n);
    for(int i=0;i<n;i++)cin >> a[i];
    for(int i=0;i<n;i++)pref[i+1]=pref[i]+a[i];

    for(int j=1;j<=k+1;j++){
        q.clear();
        pos.clear();
        for(int i=1;i<=n;i++){
            insert(pref[i-1],dp[i-1][j-1]-pref[i-1]*pref[n]);
            pos.push_front(i-1);

            dp[i][j]=query(pref[i])-pref[i]*pref[i]+pref[i]*pref[n];
            p[i][j]=pos.back();
            // for(int x=0;x<i;x++){
            //     if(dp[x][j-1]+(pref[i]-pref[x])*(pref[n]-pref[i])>=dp[i][j]){
            //         dp[i][j]=dp[x][j-1]+(pref[i]-pref[x])*(pref[n]-pref[i]);
            //         p[i][j]=x;
            //     }
            
            // cout << '(' << i << ' ' << j << ')' << dp[i][j] << ' ' << p[i][j] << ' ';
        }
        // cout << '\n';
    }

    cout << dp[n][++k] << '\n';

    int sus=p[n][k--];

    while(k>0){
        cout << sus << ' ';
        sus=p[sus][k--];
    }

    return 0;
// 7 3
// 4 1 3 4 0 2 3

}

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