제출 #563287

#	제출 시각	아이디	문제	언어	결과	실행 시간	메모리
563287		piOOE	수열 (APIO14_sequence)	C++17	100 / 100	1120 ms	88308 KiB

sequence

//#define _GLIBCXX_DEBUG

//#pragma GCC optimize("Ofast")
//#pragma GCC optimize("unroll-loops")
//#pragma GCC target("sse,sse2,sse3,ssse3,sse4,popcnt,abm,mmx,avx,tune=native")

#include <bits/stdc++.h>

using namespace std;

//#include <ext/pb_ds/assoc_container.hpp>
//
//using namespace __gnu_pbds;
//
//template<typename T>
//using ordered_set = tree<T, null_type, less < T>, rb_tree_tag, tree_order_statistics_node_update>;

//template<typename T>
//using normal_queue = priority_queue<T, vector<T>, greater<>>;

mt19937 rnd(chrono::steady_clock::now().time_since_epoch().count());

#define trace(x) cout << #x << ": " << (x) << endl;
#define all(x) begin(x), end(x)
#define rall(x) rbegin(x), rend(x)
#define uniq(x) x.resize(unique(all(x)) - begin(x))
#define sz(s) ((int) size(s))
#define pii pair<int, int>
#define mp(x, y) make_pair(x, y)
#define int128 __int128
#define pb push_back
#define popb pop_back
#define eb emplace_back
#define fi first
#define se second
#define itn int

typedef long long ll;
typedef pair<ll, ll> pll;
typedef long double ld;
typedef double db;
typedef unsigned int uint;


template<typename T>
bool ckmn(T &x, T y) {
    if (x > y) {
        x = y;
        return true;
    }
    return false;
}

template<typename T>
bool ckmx(T &x, T y) {
    if (x < y) {
        x = y;
        return true;
    }
    return false;
}

int bit(int x, int b) {
    return (x >> b) & 1;
}

int rand(int l, int r) { return (int) ((ll) rnd() % (r - l + 1)) + l; }


const ll infL = 3e18;
const int infI = 1000000000 + 7;
const int infM = 0x3f3f3f3f; //a little bigger than 1e9
const ll infML = 0x3f3f3f3f3f3f3f3fLL; //4.5e18
const int N = 100002;
const int mod = 998244353;
const ld eps = 1e-9;

struct cht_min {

    struct line {
        ll k, b;
        int id;
        double f;

        line() = default;

        line(ll a, ll B, int gg) {
            k = a, b = B, id = gg;
        }

        double intersect(line now) const {
            return (ld) (b - now.b) / (now.k - k);
        }

        ll get(ll x) const {
            return k * x + b;
        }
    };


    vector<line> d;
    int idx = 0;

    void add(line now) {
        while (d.size() > 1) {
            if (d.back().k == now.k) {
                if (d.back().b < now.b) {
                    return;
                }
            }
            if (now.intersect(d.back()) > d.back().f) break;
            d.pop_back();
        }
        if (!d.empty())
            now.f = now.intersect(d.back());
        else
            now.f = -infL;
        d.push_back(now);
    }

    pair<ll, int> get(ll x) {
        if (d.empty()) return {infL, -1};
        idx = min(idx, (int) d.size() - 1);
        while (idx < (int) d.size() - 1 && d[idx].get(x) > d[idx + 1].get(x))
            ++idx;
        return {d[idx].get(x), d[idx].id};
    }
};

int pref[N];
const int K = 201;
ll dp[2][N];
int p[K][N];

int main() {
    ios::sync_with_stdio(false);
    cin.tie(nullptr);
    int n, k;
    cin >> n >> k;
    for (int i = 1; i <= n; ++i) {
        cin >> pref[i];
        pref[i] += pref[i - 1];
    }
    //we need to minimize the squares of sums in each group
    for (int i = 1; i <= n; ++i) {
        dp[0][i] = pref[i] * 1LL * pref[i];
    }
    memset(p, -1, sizeof(p));
    for (int cut = 1; cut <= k; ++cut) {
        cht_min t;
        int nw = cut & 1;
        int pr = nw ^ 1;
        fill(dp[nw], dp[nw] + n + 1, infL);
        for (int i = 1; i <= cut; ++i) {
            t.add({-2 * pref[i], pref[i] * 1LL * pref[i] + dp[pr][i], i});
        }
        for (int i = cut + 1; i <= n; ++i) {
            auto[ff, ss] = t.get(pref[i]);
            if (ff < infL) {
                ff += pref[i] * 1LL * pref[i];
                if (ckmn(dp[nw][i], ff)) {
                    p[cut][i] = ss;
                }
                t.add({-2 * pref[i], pref[i] * 1LL * pref[i] + dp[pr][i], i});
            }
        }
    }
    vector<int> ans;
    for (int i = n, cut = k; i != -1;) {
        ans.push_back(i);
        i = p[cut--][i];
    }
    reverse(all(ans));
    ans.pop_back();
    ll val = pref[n] * 1LL * pref[n] - dp[k & 1][n];
    cout << val / 2 << '\n';
    for (int x: ans) cout << x << " ";
    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