Submission #736215

#	Time	Username	Problem	Language	Result	Execution time	Memory
736215		GrindMachine	Split the sequence (APIO14_sequence)	C++17	100 / 100	636 ms	85964 KiB

sequence

// Om Namah Shivaya

#include <bits/stdc++.h>
#include <ext/pb_ds/assoc_container.hpp>
#include <ext/pb_ds/tree_policy.hpp>

using namespace std;
using namespace __gnu_pbds;

template<typename T> using Tree = tree<T, null_type, less<T>, rb_tree_tag, tree_order_statistics_node_update>;
typedef long long int ll;
typedef long double ld;
typedef pair<int, int> pii;
typedef pair<ll, ll> pll;

#define fastio ios_base::sync_with_stdio(false); cin.tie(NULL)
#define pb push_back
#define endl '\n'
#define sz(a) a.size()
#define setbits(x) __builtin_popcountll(x)
#define ff first
#define ss second
#define conts continue
#define ceil2(x, y) ((x + y - 1) / (y))
#define all(a) a.begin(), a.end()
#define rall(a) a.rbegin(), a.rend()
#define yes cout << "Yes" << endl
#define no cout << "No" << endl

#define rep(i, n) for(int i = 0; i < n; ++i)
#define rep1(i, n) for(int i = 1; i <= n; ++i)
#define rev(i, s, e) for(int i = s; i >= e; --i)
#define trav(i, a) for(auto &i : a)

template<typename T>
void amin(T &a, T b) {
    a = min(a, b);
}

template<typename T>
void amax(T &a, T b) {
    a = max(a, b);
}

#ifdef LOCAL
#include "debug.h"
#else
#define debug(x) 42
#endif

/*

refs:
https://youtu.be/HnZKQJtGeHs (cht tutorial)

*/

const int MOD = 1e9 + 7;
const int N = 1e5 + 5;
const int inf1 = int(1e9) + 5;
const ll inf2 = ll(1e18) + 5;

ll my_ceil(ll x, ll y) {
    if (y < 0) {
        x = -x, y = -y;
    }

    ll val = -1;

    if (x >= 0) val = (x / y) + (x % y > 0);
    else val = x / y;

    return val;
}

struct Line {
    ll m, c, id, p;

    ll get(ll x) {
        return m * x + c;
    }

    ll intersect(Line li) {
        ll x = my_ceil(c - li.c, li.m - m);
        return x;
    }
};

struct CHT {
    /*

    consider:
    equal gradients
    ceil function that works for neg nums

    */

    deque<Line> dq;

    void insert(Line li) {
        while (!dq.empty()) {
            if (li.m == dq.back().m) {
                if (dq.back().c >= li.c) {
                    return;
                }
            }

            ll x = dq.back().p;
            ll old_val = dq.back().get(x);
            ll new_val = li.get(x);
            if (new_val >= old_val) {
                dq.pop_back();
            }
            else {
                break;
            }
        }

        if (dq.empty()) {
            li.p = 0;
            dq.pb(li);
        }
        else {
            li.p = li.intersect(dq.back());
            dq.pb(li);
        }
    }

    pll query(ll x) {
        while (sz(dq) > 1) {
            if (dq[1].get(x) >= dq[0].get(x)) {
                dq.pop_front();
            }
            else {
                break;
            }
        }

        return {dq[0].get(x), dq[0].id};
    }
};

void solve(int test_case)
{
    ll n, k; cin >> n >> k;
    k++;

    vector<ll> a(n + 5);
    rep1(i, n) cin >> a[i];

    vector<ll> p(n + 5);
    rep1(i, n) p[i] = p[i - 1] + a[i];

    int par[n + 5][k + 5];
    memset(par, -1, sizeof par);

    vector<ll> dp1(n + 5, -inf2), dp2(n + 5, -inf2);
    dp1[0] = 0;

    rep1(t, k) {
        fill(all(dp2), -inf2);
        CHT cht;

        rep1(i, n) {
            ll m = p[i - 1];
            ll c = (-p[i - 1] * p[n]) + dp1[i - 1];
            cht.insert({m, c, i - 1, 0});

            auto [mx, id] = cht.query(p[i]);
            dp2[i] = mx + (p[i] * p[n]) - (p[i] * p[i]);
            par[i][t] = id;
        }

        dp1 = dp2;
    }

    ll ans = dp1[n];
    cout << ans << endl;

    ll ptr = n;
    vector<ll> cuts;

    rev(t, k, 1) {
        if (ptr != n) {
            cuts.pb(ptr);
        }
        ptr = par[ptr][t];
    }

    reverse(all(cuts));
    trav(x, cuts) cout << x << " ";
    cout << endl;
}

int main()
{
    fastio;

    int t = 1;
    // cin >> t;

    rep1(i, t) {
        solve(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