제출 #396160

#	제출 시각	아이디	문제	언어	결과	실행 시간	메모리
396160		talant117408	수열 (APIO14_sequence)	C++17	100 / 100	1076 ms	81720 KiB

sequence

/*
    Code written by Talant I.D.
*/
#include <bits/stdc++.h>
#include <ext/pb_ds/assoc_container.hpp>
 
using namespace __gnu_pbds;
using namespace std;
 
typedef long long ll;
typedef pair <int, int> pii;
typedef pair <ll, ll> pll;
typedef tree <pii, null_type, less <pii>, rb_tree_tag, tree_order_statistics_node_update> ordered_set;
 
#define precision(n) fixed << setprecision(n)
#define pb push_back
#define ub upper_bound
#define lb lower_bound
#define mp make_pair
#define eps (double)1e-9
#define PI 2*acos(0.0)
#define endl "\n"
#define sz(v) int((v).size())
#define all(v) v.begin(),v.end()
#define rall(v) v.rbegin(),v.rend()
#define do_not_disturb ios::sync_with_stdio(0);cin.tie(0);cout.tie(0);
#define OK cout << "OK" << endl;
 
const int mod = 1e9+7;
 
ll mode(ll a) {
    a -= mod;
    if (a < 0) a += mod;
    return a;
}
 
ll subt(ll a, ll b) {
    return mode(mode(a)-mode(b));
}
 
ll add(ll a, ll b) {
    return mode(mode(a)+mode(b));
}
 
ll mult(ll a, ll b) {
    return mode(mode(a)*mode(b));
}
 
ll binpow(ll a, ll b) {
    ll res = 1;
    while (b) {
        if (b&1) res = mult(res, a);
        a = mult(a, a);
        b >>= 1;
    }
    return res;
}

const int N = 1e5+7;
ll dp[2][N], a[N], pref[N];
int par[203][N];

struct line {
    ll m, c;
    line(ll _m, ll _c) {
        m = _m; 
        c = _c;
    }
    ll get(ll x) {
        return m * x + c;
    } 
    ll intersect(line other) {
        return (c - other.c) / (other.m - m);
    }
};

int main() {
    do_not_disturb
    
    int n, K;
    cin >> n >> K;
    for (int i = 1; i <= n; i++) {
        cin >> a[i];
        pref[i] = pref[i-1] + a[i];
    }
    
    for (int i = 1; i <= n; i++) {
        dp[0][i] = pref[i]*(pref[n]-pref[i]);
    }
    
    for (int k = 2; k <= K; k++) {
        deque <pair <int, line>> K;
        K.push_front(mp(k-1, line(-pref[k-1], dp[0][k-1])));
        for (int i = k; i <= n; i++) {
            while (sz(K) >= 2 && K.back().second.get(pref[n]-pref[i]) <= K[sz(K)-2].second.get(pref[n]-pref[i])) {
                K.pop_back();
            }
            ll res = K.back().second.get(pref[n]-pref[i])+pref[i]*(pref[n]-pref[i]);
            dp[1][i] = res;
            par[k][i] = K.back().first;
            if (a[i]) {
                line cur = line(-pref[i], dp[0][i]);
                while (sz(K) >= 2 && (cur.intersect(K[0].second) >= K[0].second.intersect(K[1].second))) {
                    K.pop_front();
                }
                K.push_front(mp(i, cur));
            }
        }
        for (int i = 1; i <= n; i++) {
            dp[0][i] = dp[1][i];
        }
    }
    
    ll mx = -1;
    int ind = 0;
    for (int i = 1; i <= n; i++) {
        if (dp[0][i] > mx) {
            mx = dp[0][i];
            ind = i;
        }
    }
    
    printf("%lld\n", mx);
    vector <int> v;
    for (int i = 0; i < K; i++) {
        v.pb(ind);
        ind = par[K - i][ind];
    }
    reverse(all(v));
    for (auto to : v) printf("%d ", to);
    
    return 0;
}
/*
7 3
4 1 3 4 0 2 3
*/

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