제출 #562824

#	제출 시각	아이디	문제	언어	결과	실행 시간	메모리
562824		Ooops_sorry	수열 (APIO14_sequence)	C++14	71 / 100	771 ms	131072 KiB

sequence

#include<bits/stdc++.h>

using namespace std;

#define pb push_back
#define ld long double
#define ll long long

mt19937 rnd(51);

const ll INF = 1e18, K = 210;

struct line {
    ll k, b;
    int i;
};

vector<line> q[K];

ld intersection(line a, line b) {
    return (ld)(b.b - a.b) / (a.k - b.k);
}

void add(int i, line a) {
    while (q[i].size() > 1) {
        int sz = q[i].size();
        if (q[i].back().k == a.k) {
            if (a.b > q[i].back().b) {
                q[i].pop_back();
            } else {
                return;
            }
        }
        if (intersection(q[i][sz - 2], a) < intersection(q[i][sz - 2], q[i][sz - 1])) {
            q[i].pop_back();
        } else {
            break;
        }
    }
    if (q[i].size() == 1 && q[i].back().k == a.k) {
        if (a.b > q[i].back().b) {
            q[i].pop_back();
        } else {
            return;
        }
    }
    q[i].pb(a);
}

pair<ll, int> get(int i, ll x) {
    int l = -1, r = (int)(q[i].size()) - 1;
    if (q[i].size() == 0) return {-INF, -1};
    while (r - l > 1) {
        int mid = (r + l) / 2;
        if (intersection(q[i][mid], q[i][mid + 1]) <= x) {
            l = mid;
        } else {
            r = mid;
        }
    }
    return {x * q[i][r].k + q[i][r].b, q[i][r].i};
}

signed main() {
#ifdef LOCAL
    freopen("input.txt", "r", stdin);
#endif // LCOAL
    ios_base::sync_with_stdio(0);
    cin.tie(0);
    int n, k;
    cin >> n >> k;
    vector<ll> a(n), pr(n);
    vector<vector<int>> par(n + 1, vector<int>(k + 1));
    for (int i = 0; i < n; i++) {
        cin >> a[i];
    }
    reverse(a.begin(), a.end());
    for (int i = 0; i < n; i++) {
        pr[i] = a[i];
        if (i > 0) {
            pr[i] += pr[i - 1];
        }
    }
    ll ans = -1;
    for (int i = 0; i < n; ++i) {
        for (int f = k; f >= 1; --f) {
            pair<ll, int> p = get(f - 1, pr[i]);
            if (i == n - 1 && f == k) {
                ans = p.first;
            }
            par[i][f] = p.second;
            add(f, {pr[i], p.first - pr[i] * pr[i], i});
        }
        add(0, {pr[i], 0 - pr[i] * pr[i], i});
    }
    cout << ans << endl;
    vector<int> res;
    int i = n - 1, j = k, sum = 0;
    while (j > 0) {
        res.pb(i - par[i][j]);
        i = par[i][j];
        j--;
        sum += res.back();
    }
    int now = 0;
    for (auto to : res) {
        cout << (now += to) << ' ';
    }
    cout << endl;
    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