Submission #1088369

#TimeUsernameProblemLanguageResultExecution timeMemory
1088369rlx0090Split the sequence (APIO14_sequence)C++17
0 / 100
126 ms16056 KiB
#include <iostream> #include <string> #include <vector> #include <algorithm> #include <cmath> #include <limits> #include <queue> #include <string.h> #include <stack> #include <limits> using namespace std; int N, K; int cache[205][100005], p[100005]; long long dp[2][100005]; long long minf = numeric_limits<long long>::min(); struct seg { long long k = 0, b = minf; int label = -1; seg() {} seg(long long _k, long long _b, int _label) :k(_k), b(_b), label(_label) {} long long f(long long x) { return k * x + b; } }; bool which_is_big(seg a, seg b, int x) { long long l = a.f(x), r = b.f(x); if(l == r) return a.label < b.label; return l > r; } struct Node { seg s; int l = -1, r = -1; Node() {} }; int pv = 2; Node tree[400005]; Node null_Node; void add(int n, seg ns, int nl, int nr) { if(ns.b == minf) return; int m = (nl + nr) / 2; bool lef = which_is_big(ns, tree[n].s, nl); bool mid = which_is_big(ns, tree[n].s, m); if(mid) swap(tree[n].s, ns); if(nl == nr) return; else if(lef != mid) { if(tree[n].l == -1) tree[n].l = pv++; add(tree[n].l, ns, nl, m); } else { if(tree[n].r == -1) tree[n].r = pv++; add(tree[n].r, ns, m + 1, nr); } } seg query(int n, int x, int nl, int nr) { if(n == -1) return seg(); if(nl == nr) return tree[n].s; int m = (nl + nr) / 2; if(x <= m){ seg c = query(tree[n].l, x, nl, m); if(which_is_big(c, tree[n].s, x)) return c; return tree[n].s; } else { seg c = query(tree[n].r, x, m + 1, nr); if(which_is_big(c, tree[n].s, x)) return c; return tree[n].s; } } int main() { ios::sync_with_stdio(false); cin.tie(NULL); cout.tie(NULL); cin >> N >> K; long long q; for(int i = 1; i <= N; ++i) { cin >> q; p[i] = p[i - 1] + q; } int before = 0, cur = 0; for(int k = 1; k <= K; ++k) { before = k % 2; cur = (k + 1) % 2; for(int i = N - k - 1; i >= 0; --i) { add(1, seg(p[i + 1], - p[i + 1] * p[i + 1] + p[N] * p[i + 1] + dp[before][i + 1], i + 1), 0, 1e9); seg c = query(1, p[i], 0, 1e9); dp[cur][i] = c.f(p[i]) - p[N] * p[i]; cache[k][i] = c.label; } for(int i = 1; i <= 4 * N; ++i) tree[i] = null_Node; pv = 2; } cout << dp[cur][0] << '\n'; int s = 0; for(int k = K; k >= 1; --k) { cout << cache[k][s] << ' '; s = cache[k][s]; } return 0; }
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