Submission #329236

#	Time	Username	Problem	Language	Result	Execution time	Memory
329236		jovan_b	Split the sequence (APIO14_sequence)	C++17	71 / 100	2080 ms	25568 KiB

sequence

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

typedef long long ll;

ll pre[100005];
ll a[100005];
ll dp[100005];
ll dpp[100005];
int prosli[205][100005];

struct line{
    ll a, b;
    int ind;
    int level;
    ll eval(ll x){
        return a*x+b;
    }
};

line seg[400005];

void upd(int node, int l, int r, line g){
    int mid = (l+r)/2;
    if((g.eval(pre[l]) >= seg[node].eval(pre[l]) && g.eval(pre[r]) >= seg[node].eval(pre[r])) || seg[node].level != g.level){
        seg[node] = g;
        return;
    }
    if(l == r) return;
    if(g.eval(pre[l]) > seg[node].eval(pre[l]) || g.eval(pre[mid]) > seg[node].eval(pre[mid])) upd(node*2, l, mid, g);
    if(g.eval(pre[r]) > seg[node].eval(pre[r]) || g.eval(pre[mid+1]) > seg[node].eval(pre[mid+1])) upd(node*2+1, mid+1, r, g);
}

pair <int, ll> query(int node, int l, int r, int x, int lvl){
    pair <int, ll> res1 = {seg[node].ind, seg[node].eval(pre[x])};
    if(lvl != seg[node].level) return {0, 0};
    if(l == r) return res1;
    int mid = (l+r)/2;
    pair <int, ll> res2;
    if(x <= mid) res2 = query(node*2, l, mid, x, lvl);
    else res2 = query(node*2+1, mid+1, r, x, lvl);
    if(res1.second >= res2.second) return res1;
    else return res2;
}

void init(int node, int l, int r){
    if(l == r){
        seg[node].a = -100000;
        seg[node].b = -1000000000000000LL;
        seg[node].ind = 0;
        seg[node].level = 0;
        return;
    }
    int mid = (l+r)/2;
    init(node*2, l, mid);
    init(node*2+1, mid+1, r);
}

int main(){
    ios_base::sync_with_stdio(false), cin.tie(0);

    int n, k;
    cin >> n >> k;
    for(int i=1; i<=n; i++){
        cin >> a[i];
        pre[i] = pre[i-1] + a[i];
    }
    init(1, 1, n);
    for(int j=1; j<=k; j++){
        for(int i=1; i<=n; i++){
            pair <int, ll> d = query(1, 1, n, i, j);
            dp[i] = d.second;
            prosli[j][i] = d.first;
            line nline;
            nline.a = pre[i];
            nline.b = -pre[i]*pre[i] + dpp[i];
            nline.ind = i;
            nline.level = j;
            upd(1, 1, n, nline);
        }
        for(int i=1; i<=n; i++){
            dpp[i] = dp[i];
        }
    }
    cout << dp[n] << "\n";
    int t = n;
    for(int i=k; i>=1; i--){
        cout << prosli[i][t] << " ";
        t = prosli[i][t];
    }
    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: 0 / 29