Submission #655290

#	Time	Username	Problem	Language	Result	Execution time	Memory
655290		benjaminkleyn	Split the sequence (APIO14_sequence)	C++17	71 / 100	2101 ms	89420 KiB

sequence

#pragma GCC optimize("Ofast")
#pragma GCC optimize("unroll-loops")
#pragma GCC target("avx,avx2,fma")
#include <bits/stdc++.h>
using namespace std;
typedef long long ll;
 
struct Line {
    mutable ll k, m, p, idx;
    bool operator<(const Line& o) const { return k < o.k; }
    bool operator<(ll x) const { return p < x; }
};
 
struct LineContainer : multiset<Line, less<>> {
    static const ll inf = LLONG_MAX;
    ll div(ll a, ll b) {
        return a / b - ((a ^ b) < 0 && a % b); }
    bool isect(iterator x, iterator y) {
        if (y == end()) return x->p = inf, 0;
        if (x->k == y->k) x->p = x->m > y->m ? inf : -inf;
        else x->p = div(y->m - x->m, x->k - y->k);
        return x->p >= y->p;
    }
    void add(ll k, ll m, int idx) {
        auto z = insert({k, m, 0, idx}), y = z++, x = y;
        while (isect(y, z)) z = erase(z);
        if (x != begin() && isect(--x, y)) isect(x, y = erase(y));
        while ((y = x) != begin() && (--x)->p >= y->p)
            isect(x, erase(y));
    }
    pair<ll,int> query(ll x) {
        assert(!empty());
        auto l = *lower_bound(x);
        return {l.k * x + l.m, l.idx};
    }
};
 
int N, K, a[100001];
ll pref[100001] = {0};
ll dp[100001][2] = {0};
int last_split[100001][201];
 
int main()
{
    ios_base::sync_with_stdio(false);
    cin.tie(NULL);

    cin >> N >> K;
    for (int i = 1; i <= N; i++)
    {
         cin >> a[i];
         pref[i] = pref[i-1] + a[i];
    }
 
    LineContainer cht;
    for (int k = 1; k <= K; k++)
    {
        cht.clear();
        for (int i = k + 1; i <= N; i++)
        {
            dp[i][1] = 0;
            cht.add(pref[i-1], dp[i-1][0] - pref[i-1] * pref[i-1], i-1);
            auto [val, idx] = cht.query(pref[i]);
            dp[i][1] = val;
            last_split[i][k] = idx;
        }
        for (int i = 1; i <= N; i++)
            dp[i][0] = dp[i][1];
    }
 
    cout << dp[N][0] << '\n';
 
    int i = N;
    do
    {
        i = last_split[i][K];
        cout << i << ' ';
    } while (--K);
    cout << '\n';
 
    return 0;
}

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