Submission #531459

#	Time	Username	Problem	Language	Result	Execution time	Memory
531459		Soumya1	Split the sequence (APIO14_sequence)	C++17	100 / 100	795 ms	83464 KiB

sequence

#include <bits/stdc++.h>
#ifdef __LOCAL__
#include <debug_local.h>
#endif
using namespace std;
using ll = long long;
struct Line {
  ll m = 0, c = 0, id = 0;
  Line(ll _m, ll _c, ll _id) : m(_m), c(_c), id(_id) { }
  ll get(ll x) { return m * x + c; }
  double intersection(Line o) { return (double) (o.c - c) / (double) (m - o.m); }
  bool good(Line l2, Line l3) { return __int128(l2.c - c) * (l2.m - l3.m) < __int128(l3.c - l2.c) * (m - l2.m); }
};
struct CHT {
  deque<Line> hull;
  void insert(Line l) {
    while (hull.size() > 1) {
      auto last = hull.end()[-1];
      auto p_last = hull.end()[-2];
      if (p_last.good(last, l)) break;
      hull.pop_back();
    }
    hull.push_back(l);
  }
  pair<ll, ll> query(ll x) {
    while (hull.size() > 1 && hull[0].get(x) < hull[1].get(x)) {
      hull.pop_front();
    }
    return {hull[0].get(x), hull[0].id};
  }
};
int par[100002][202];
void testCase() {
  int n, K;
  cin >> n >> K;
  vector<ll> a(n + 1);
  for (int i = 1; i <= n; i++) {
    cin >> a[i];
    a[i] += a[i - 1];
  }
  vector<ll> dp(n + 1);
  for (int k = 1; k <= K + 1; k++) {
    vector<ll> new_dp(n + 1);
    CHT cht;
    cht.insert(Line(a[k - 1], dp[k - 1] - a[k - 1] * a[n], k - 1));
    for (int i = k; i <= n; i++) {
      auto [x, id] = cht.query(a[i]);
      par[i][k] = id;
      x += a[n] * a[i] - a[i] * a[i];
      new_dp[i] = x;
      cht.insert(Line(a[i], dp[i] - a[n] * a[i], i));
    }
    // debug(new_dp);
    if (k == K + 1) {
      cout << new_dp[n] << "\n";
    }
    dp = new_dp;
  }
  vector<int> ans;
  {
    int cur = n;
    int k = K + 1;
    while (cur) {
      ans.push_back(cur);
      cur = par[cur][k];
      k--;
    }
  }
  reverse(ans.begin(), ans.end());
  ans.pop_back();
  for (int i : ans) cout << i << " ";
  cout << "\n";
}
int main() {
  ios::sync_with_stdio(false);
  cin.tie(nullptr);
  testCase();
  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