Submission #981782

#TimeUsernameProblemLanguageResultExecution timeMemory
981782Halym2007Split the sequence (APIO14_sequence)C++17
100 / 100
694 ms84564 KiB
#include <bits/stdc++.h> using namespace std; const int N = 1e5 + 5; #define ll long long #define sz size() #define pb push_back ll p[N], a[N], n, k, dp[N]; int par[N][205]; //CHT means convex hull trick struct CHT { ll m, c; int idx; ll get (ll x) { return m * x + c; } long double intersect (CHT x) { return (long double)(x.c - c) / (long double)(m - x.m); } }; int main () { // freopen ("input.txt", "r", stdin); cin >> n >> k; for (int i = 1; i <= n; ++i) { cin >> a[i]; p[i] = p[i - 1] + a[i]; } for (int i = 1; i <= k; ++i) { deque <CHT> d; d.pb ({p[i], dp[i] - p[i] * p[i], i}); for (int j = i + 1; j <= n; ++j) { while ((int)d.sz >= 2 and d[0].get (p[j]) <= d[1].get (p[j])) d.pop_front(); ll aa = d[0].get (p[j]); CHT tr = {p[j], dp[j] - p[j] * p[j], j}; par[j][i] = d[0].idx; while ((int)d.sz >= 2 and tr.intersect(d[(int)d.sz - 2]) <= d.back().intersect(d[(int)d.sz - 2])) d.pop_back(); d.pb (tr); dp[j] = aa; } } cout << dp[n] << "\n"; int idx = par[n][k]; int val = k; while (idx) { val--; cout << idx << " "; idx = par[idx][val]; } }
#Verdict Execution timeMemoryGrader output
Fetching results...
#Verdict Execution timeMemoryGrader output
Fetching results...
#Verdict Execution timeMemoryGrader output
Fetching results...
#Verdict Execution timeMemoryGrader output
Fetching results...
#Verdict Execution timeMemoryGrader output
Fetching results...
#Verdict Execution timeMemoryGrader output
Fetching results...