제출 #142510

#	제출 시각	아이디	문제	언어	결과	실행 시간	메모리
142510		meatrow	수열 (APIO14_sequence)	C++17	0 / 100	211 ms	7416 KiB

sequence

//#pragma GCC optimize("O3")
//#pragma GCC target("sse,sse2,sse3,ssse3,sse4,popcnt,abm,mmx,tune=native")
//#pragma GCC optimize ("unroll-loops")
#include <bits/stdc++.h>
 
using namespace std;
 
using ll = long long;
using ld = long double;
 
const int mod = 1e9 + 7;
const int N = 1e5 + 5;
const int K = 205;


struct Line {
    ll k, b;
    int pos;
    ll eval(ld x) {
        return k * x + b;
    }
};
 
inline ld isec(Line &p, Line &q) {
    return ((ld)p.b - q.b) / (q.k - p.k);
}
 
struct CHT {
    int s;
    Line l[N];
    //int pt;
    void clr() {
        s = 0;// pt = 0;
    }
    void add(Line a) {
        while(s >= 2 && isec(l[s - 2], l[s - 1]) >= isec(l[s - 1], a)) s--;//, pt = min(pt, s - 1);
        l[s++] = a;
    }
    pair<ll, int> get(ll x) {
        int lf = 1, rg = s - 1;
        if(s == 1 || x <= isec(l[0], l[1])) lf = 0;
        else while(lf < rg) {
            int md = (lf + rg + 1) >> 1;
            if(isec(l[md - 1], l[md]) <= x) lf = md;
            else rg = md - 1;
        }
        return { l[lf].eval(x), l[lf].pos };
    }
//
//    ll get(ll x) {
//        while(pt != s - 1 && isec(l[pt], l[pt + 1]) <= x) pt++;
//        return l[pt].eval(x);
//    }
};


int a[N];
ll pref[N];
ll dp[K][N];
int way[K][N];
CHT cht;

int main() {
    ios::sync_with_stdio(false);
    cin.tie(nullptr);
    cout.tie(nullptr);
    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 = 2; i <= n; i++) {
        int pos = upper_bound(pref + 1, pref + i + 1, pref[i] / 2) - pref - 1;
        dp[1][i] = (pref[i] - pref[pos]) * pref[pos];
        way[1][i] = pos;
    }
    for (int i = 2; i <= k; i++) {
        cht.clr();
        cht.add({ 0, 0, 0 });
        for (int j = 1; j <= n; j++) {
            auto p = cht.get(pref[j]);
            dp[i][j] = p.first;
            way[i][j] = p.second;
            cht.add({ pref[j], dp[i - 1][j] - pref[j] * pref[j], j });
        }
    }
    cout << dp[k][n] << '\n';
    int kek = n;
    vector<int> ans;
    for (int i = k; i > 0; i--) {
        ans.push_back(way[i][kek]);
        kek = way[i][kek];
    }
    reverse(ans.begin(), ans.end());
    for (int i : ans) {
        cout << i << ' ';
    }
    return 0;
}

• Subtask 1: 0 / 11
• Subtask 2: 0 / 11
• Subtask 3: 0 / 11
• Subtask 4: 0 / 17
• Subtask 5: 0 / 21
• Subtask 6: 0 / 29