Submission #556682

#	Time	Username	Problem	Language	Result	Execution time	Memory
556682		timreizin	Split the sequence (APIO14_sequence)	C++17	100 / 100	746 ms	86876 KiB

sequence

#pragma GCC optimize("Ofast,unroll-loops")
#pragma GCC target("avx2,bmi,bmi2,popcnt,lzcnt")

#include <iostream>
#include <vector>
#include <algorithm>
#include <deque>
#include <set>
#include <string>
#include <numeric>
#include <cmath>
#include <cassert>
#include <array>
#include <numeric>

using namespace std;

using ll = long long;

const ll INF = 1e18;

struct Line
{
    ll k, b;
    int who;
    
    Line(ll k = 0, ll b = 0, int who = 0) : k(k), b(b), who(who)
    {}
    
    ll operator()(ll x)
    {
        return k * x + b;
    }
    
};

bool check(Line a, Line b, Line c)
{
    if (b.k == c.k) return b.b >= c.b;
    return (__int128)(a.b - c.b) * (b.k - a.k) <= (__int128)(a.b - b.b) * (c.k - a.k);
}

int main()
{
    cin.tie(0)->sync_with_stdio(0);
    int n, k;
    cin >> n >> k;
    ++k;
    vector<ll> a(n);
    for (ll &i : a) cin >> i;
    array<vector<ll>, 2> dp;
    dp[0] = dp[1] = vector<ll>(n + 1, INF);
    dp[0][0] = 0;
    ll res = accumulate(a.begin(), a.end(), 0ll);
    res = res * res;
    partial_sum(a.begin(), a.end(), a.begin());
    a.insert(a.begin(), 0);
    vector<vector<int>> parent(n + 1, vector<int>(k, -1));
    for (int i = 0; i < k; ++i)
    {
        deque<Line> dq;
        dq.push_back({-2 * a[i], dp[0][i] + a[i] * a[i], i});
        for (int j = i + 1; j <= n; ++j)
        {
            while (dq.size() >= 2 && dq[0](a[j]) > dq[1](a[j])) dq.pop_front();
            dp[1][j] = dq.front()(a[j]) + a[j] * a[j];
            parent[j][i] = dq.front().who;
            Line add{-2 * a[j], dp[0][j] + a[j] * a[j], j};
            while (dq.size() >= 2 && check(dq[(int)dq.size() - 2], dq.back(), add)) dq.pop_back();
            if (dq.back().k != add.k) dq.push_back(add);
        }
        dp[0] = dp[1];
        dp[1] = vector<ll>(n + 1, INF);
    }
    vector<int> path(k - 1);
    for (int i = k - 1, v = n; i > 0; v = parent[v][i], --i) path[k - i - 1] = parent[v][i];
    cout << (res - dp[0].back()) / 2 << '\n';
    reverse(path.begin(), path.end());
    for (int i : path) cout << i << ' ';
    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