結果

問題 No.919 You Are A Project Manager
ユーザー hitonanode
提出日時 2019-10-25 23:10:40
言語 C++14
(gcc 13.3.0 + boost 1.87.0)
結果
AC  
実行時間 994 ms / 3,000 ms
コード長 6,323 bytes
コンパイル時間 2,499 ms
コンパイル使用メモリ 194,272 KB
実行使用メモリ 6,948 KB
最終ジャッジ日時 2024-06-27 17:26:48
合計ジャッジ時間 26,959 ms
ジャッジサーバーID
(参考情報)
judge2 / judge1
このコードへのチャレンジ
(要ログイン)
ファイルパターン 結果
sample AC * 3
other AC * 55
権限があれば一括ダウンロードができます

ソースコード

diff #
プレゼンテーションモードにする

#include <bits/stdc++.h>
using namespace std;
using lint = long long int;
using pint = pair<int, int>;
using plint = pair<lint, lint>;
struct fast_ios { fast_ios(){ cin.tie(0); ios::sync_with_stdio(false); cout << fixed << setprecision(20); }; } fast_ios_;
#define ALL(x) (x).begin(), (x).end()
#define SZ(x) ((lint)(x).size())
#define POW2(n) (1LL << (n))
#define FOR(i, begin, end) for(int i=(begin),i##_end_=(end);i<i##_end_;i++)
#define IFOR(i, begin, end) for(int i=(end)-1,i##_begin_=(begin);i>=i##_begin_;i--)
#define REP(i, n) FOR(i,0,n)
#define IREP(i, n) IFOR(i,0,n)
template<typename T> void ndarray(vector<T> &vec, int len) { vec.resize(len); }
template<typename T, typename... Args> void ndarray(vector<T> &vec, int len, Args... args) { vec.resize(len); for (auto &v : vec) ndarray(v, args
    ...); }
template<typename T> bool mmax(T &m, const T q) { if (m < q) {m = q; return true;} else return false; }
template<typename T> bool mmin(T &m, const T q) { if (m > q) {m = q; return true;} else return false; }
template<typename T1, typename T2> pair<T1, T2> operator+(const pair<T1, T2> &l, const pair<T1, T2> &r) { return make_pair(l.first + r.first, l
    .second + r.second); }
template<typename T1, typename T2> pair<T1, T2> operator-(const pair<T1, T2> &l, const pair<T1, T2> &r) { return make_pair(l.first - r.first, l
    .second - r.second); }
template<typename T> istream &operator>>(istream &is, vector<T> &vec){ for (auto &v : vec) is >> v; return is; }
///// This part below is only for debug, not used /////
template<typename T> ostream &operator<<(ostream &os, const vector<T> &vec){ os << "["; for (auto v : vec) os << v << ","; os << "]"; return os; }
template<typename T> ostream &operator<<(ostream &os, const deque<T> &vec){ os << "deq["; for (auto v : vec) os << v << ","; os << "]"; return os; }
template<typename T> ostream &operator<<(ostream &os, const set<T> &vec){ os << "{"; for (auto v : vec) os << v << ","; os << "}"; return os; }
template<typename T> ostream &operator<<(ostream &os, const unordered_set<T> &vec){ os << "{"; for (auto v : vec) os << v << ","; os << "}"; return
    os; }
template<typename T> ostream &operator<<(ostream &os, const multiset<T> &vec){ os << "{"; for (auto v : vec) os << v << ","; os << "}"; return os; }
template<typename T> ostream &operator<<(ostream &os, const unordered_multiset<T> &vec){ os << "{"; for (auto v : vec) os << v << ","; os << "}";
    return os; }
template<typename T1, typename T2> ostream &operator<<(ostream &os, const pair<T1, T2> &pa){ os << "(" << pa.first << "," << pa.second << ")"; return
    os; }
template<typename TK, typename TV> ostream &operator<<(ostream &os, const map<TK, TV> &mp){ os << "{"; for (auto v : mp) os << v.first << "=>" << v
    .second << ","; os << "}"; return os; }
template<typename TK, typename TV> ostream &operator<<(ostream &os, const unordered_map<TK, TV> &mp){ os << "{"; for (auto v : mp) os << v.first << "
    =>" << v.second << ","; os << "}"; return os; }
#define dbg(x) cerr << #x << " = " << (x) << " (L" << __LINE__ << ") " << __FILE__ << endl;
///// END /////
/*
#include <ext/pb_ds/assoc_container.hpp>
#include <ext/pb_ds/tree_policy.hpp>
#include <ext/pb_ds/tag_and_trait.hpp>
using namespace __gnu_pbds; // find_by_order(), order_of_key()
template<typename TK> using pbds_set = tree<TK, null_type, less<TK>, rb_tree_tag, tree_order_statistics_node_update>;
template<typename TK, typename TV> using pbds_map = tree<TK, TV, less<TK>, rb_tree_tag, tree_order_statistics_node_update>;
*/
using T = vector<lint>;
struct SegTree
{
int N;
int head;
vector<T> x;
T defaultT;
using func = function<T(const T&, const T&)>;
func merger;
lint _get(int begin, int end, int pos, int l, int r, lint q) const
{
if (r <= begin or l >= end) return 0;
if (l >= begin and r <= end)
{
return upper_bound(ALL(x[pos]), q) - x[pos].begin();
}
return _get(begin, end, 2 * pos + 1, l, (l + r) / 2, q) + _get(begin, end, 2 * pos + 2, (l + r) / 2, r, q);
}
SegTree(int N, T defaultT, func merger) : N(N), defaultT(defaultT), merger(merger)
{
int N_tmp = 1; while (N_tmp < N) N_tmp <<= 1;
x.assign(N_tmp*2, defaultT), head = N_tmp - 1;
}
SegTree(const vector<T> &vals, T defaultT, func merger) : N(vals.size()), defaultT(defaultT), merger(merger)
{
int N_tmp = 1; while (N_tmp < N) N_tmp <<= 1;
x.assign(N_tmp*2, defaultT), head = N_tmp - 1;
copy(vals.begin(), vals.end(), x.begin() + head);
IREP(i, head) x[i] = merger(x[i * 2 + 1], x[i * 2 + 2]);
}
SegTree() : SegTree(0, T(), [](T, T) { return T(); }) {}
void build(const vector<T> &vals) { copy(vals.begin(), vals.end(), x.begin() + head); IREP(i, head) x[i] = merger(x[i * 2 + 1], x[i * 2 + 2]); }
void update(int pos, T val) { pos += head, x[pos] = val; while (pos) pos = (pos - 1) / 2, x[pos] = merger(x[pos*2+1], x[pos*2+2]); }
lint get(int begin, int end, lint q) const { return _get(begin, end, 0, 0, (int)x.size() / 2, q); }
// friend ostream &operator<<(ostream &os, const SegTree &s) { os << "["; REP(i, s.N) os << s.get(i, i + 1) << ","; os << "]"; return os; }
lint median(int l, int r)
{
int lo = -1e9 - 100, hi = 1e9 + 100;
while (hi - lo > 1)
{
int c = (lo + hi) / 2;
lint n = get(l, r, c); // c
if (n + (r - l) / 2 >= r - l) hi = c;
else lo = c;
}
return hi;
}
};
int N;
vector<vector<lint>> A;
lint ret = 0;
int main()
{
cin >> N;
A.resize(N);
REP(i, N)
{
lint a;
cin >> a;
A[i] = vector<lint>(1, a);
}
SegTree st(A, vector<lint>{}, [](T l, const T &r) {
l.insert(l.end(), ALL(r));
sort(ALL(l));
return l;
});
FOR(k, 1, N + 1)
{
int nmax = N / k;
vector<lint> l(nmax + 1), r(nmax + 1);
REP(i, nmax)
{
l[i + 1] = st.median(i * k, (i + 1) * k);
r[i + 1] = st.median(N - (i + 1) * k, N - i * k);
}
REP(i, nmax) l[i + 1] += l[i], r[i + 1] += r[i];
REP(i, nmax) mmax(l[i + 1], l[i]), mmax(r[i + 1], r[i]);
REP(i, nmax + 1) mmax(ret, (l[i] + r[nmax - i]) * k);
}
cout << ret << endl;
}
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