제출 #966193

#	제출 시각	아이디	문제	언어	결과	실행 시간	메모리
966193		Pring	수열 (APIO14_sequence)	C++17	71 / 100	66 ms	131072 KiB

sequence

#include <bits/stdc++.h>
using namespace std;

#ifdef MIKU
string dbmc = "\033[1;38;2;57;197;187m", dbrs = "\033[0m";
#define debug(x...) cout << dbmc << "[" << #x << "]: ", dout(x)
void dout() { cout << dbrs << endl; }
template <typename T, typename ...U>
void dout(T t, U ...u) { cout << t << (sizeof...(u) ? ", " : ""); dout(u...); }
#else
#define debug(...) 39
#endif

#define fs first
#define sc second
#define mp make_pair
#define FOR(i, j, k) for (int i = j, Z = k; i < Z; i++)
using ll = long long;
typedef pair<int, int> pii;

const int MXN = 100005, MXM = 205;
const ll INF = 3.9e18;
int n, k, a[MXN];
ll pre[MXN];
ll dp[MXN][MXM];
int K[MXN][MXM];

struct MK {
    ll m, k, id;
    MK() {}
    MK(ll _m, ll _k, ll _id) : m(_m), k(_k), id(_id) {}
    ll operator()(int x) {
        return m * x + k;
    }
    double operator*(MK o) {
        return (double) (o.k - k) / (m - o.m);
    }
};

deque<MK> dq;

void POP(int x) {
    while (dq.size() >= 2) {
        MK L0 = dq[0], L1 = dq[1];
        if (L0(x) < L1(x)) break;
        dq.pop_front();
    }
}

void PUSH(MK L) {
    while (dq.size() >= 2) {
        MK L_1 = dq[dq.size() - 1], L_2 = dq[dq.size() - 2];
        if (L_1.m == L.m) {
            if (L_1.k < L.k) return;
            dq.pop_back();
            continue;
        }
        if (L_1 * L_2 < L * L_2) break;
        dq.pop_back();
    }
    dq.push_back(L);
}

void DP(int id) {
    dq.push_back(MK(0, INF, 0));
    FOR(i, 1, n + 1) {
        POP(pre[i]);
        dp[i][id] = pre[i] * pre[i] + dq.front()(pre[i]);
        K[i][id] = dq.front().id;
        PUSH(MK(-2 * pre[i], dp[i][id - 1] + pre[i] * pre[i], i));
    }
    dq.clear();
}

vector<int> BT() {
    int now = n;
    vector<int> v;
    for (int i = k + 1; i > 1; i--) {
        v.push_back(K[now][i]);
        now = K[now][i];
    }
    return v;
}

void miku() {
    cin >> n >> k;
    FOR(i, 1, n + 1) cin >> a[i];
    FOR(i, 1, n + 1) pre[i] = pre[i - 1] + a[i];
    FOR(i, 1, n + 1) dp[i][1] = pre[i] * pre[i];
    FOR(i, 2, k + 2) DP(i);
    // FOR(id, 1, k + 2) {
    //     FOR(i, 1, n + 1) cout << dp[i][id] << ' ';
    //     cout << '\n';
    // }
    // cout << '\n';
    cout << (pre[n] * pre[n] - dp[n][k + 1]) / 2 << '\n';
    vector<int> ans = BT();
    for (auto &i : ans) cout << i << ' ';
    cout << '\n';
}

int32_t main() {
    cin.tie(0) -> sync_with_stdio(false);
    cin.exceptions(cin.failbit);
    miku();
    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