제출 #396113

#제출 시각아이디문제언어결과실행 시간메모리
396113talant117408수열 (APIO14_sequence)C++17
0 / 100
356 ms17428 KiB
/* 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], 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[1][i] = pref[i]*(pref[n]-pref[i]); } for (int k = 2; k <= K; k++) { int type = k&1; deque <pair <int, line>> K; vector <int> inds(n); iota(all(inds), 0); auto cmp = [&K](int i, int x) { return K[i].second.intersect(K[i + 1].second) < x; }; K.push_front(mp(k-1, line(-pref[k-1], dp[1-type][k-1]))); for (int i = k; i < n; i++) { int it = *lb(inds.begin(), inds.begin()+sz(K)-1, pref[n]-pref[i], cmp); ll res = K[it].second.get(pref[n]-pref[i])+pref[i]*(pref[n]-pref[i]); dp[type][i] = res; par[k][i] = K[it].first; line cur = line(-pref[i], dp[1-type][i]); while (sz(K) >= 2 && ((cur.m == K[0].second.m && cur.c == K[0].second.c) || cur.intersect(K[0].second) >= K[0].second.intersect(K[1].second))) { K.pop_front(); } K.push_front(mp(i, cur)); } } int mx = -1, ind = 0; for (int i = 1; i <= n; i++) { if (dp[K&1][i] > mx) { mx = dp[K&1][i]; ind = i; } } cout << mx << endl; vector <int> v; while (ind) { v.pb(ind); ind = par[K][ind]; K--; } reverse(all(v)); for (auto to : v) cout << to << ' '; return 0; }
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