結果

問題 No.68 よくある棒を切る問題 (2)
ユーザー fumiphysfumiphys
提出日時 2019-08-25 21:34:35
言語 C++14
(gcc 12.3.0 + boost 1.83.0)
結果
AC  
実行時間 434 ms / 5,000 ms
コード長 3,566 bytes
コンパイル時間 1,931 ms
コンパイル使用メモリ 178,664 KB
実行使用メモリ 10,368 KB
最終ジャッジ日時 2024-11-07 01:27:21
合計ジャッジ時間 14,613 ms
ジャッジサーバーID
（参考情報） 		judge4 / judge3
このコードへのチャレンジ
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テストケース

テストケース表示
入力 結果 実行時間
実行使用メモリ
testcase_00 AC 426 ms
10,368 KB
testcase_01 AC 431 ms
10,240 KB
testcase_02 AC 434 ms
10,236 KB
testcase_03 AC 415 ms
10,112 KB
testcase_04 AC 418 ms
10,324 KB
testcase_05 AC 407 ms
10,240 KB
testcase_06 AC 424 ms
9,724 KB
testcase_07 AC 426 ms
10,164 KB
testcase_08 AC 400 ms
9,644 KB
testcase_09 AC 411 ms
9,856 KB
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 # 

// includes
#include <bits/stdc++.h>
using namespace std;

// macros
#define pb emplace_back
#define mk make_pair
#define FOR(i, a, b) for(int i=(a);i<(b);++i)
#define rep(i, n) FOR(i, 0, n)
#define rrep(i, n) for(int i=((int)(n)-1);i>=0;i--)
#define irep(itr, st) for(auto itr = (st).begin(); itr != (st).end(); ++itr)
#define irrep(itr, st) for(auto itr = (st).rbegin(); itr != (st).rend(); ++itr)
#define all(x) (x).begin(),(x).end()
#define sz(x) ((int)(x).size())
#define UNIQUE(v) v.erase(unique(v.begin(), v.end()), v.end())
#define bit(n) (1LL<<(n))
// functions
template <class T>bool chmax(T &a, const T &b){if(a < b){a = b; return 1;} return 0;}
template <class T>bool chmin(T &a, const T &b){if(a > b){a = b; return 1;} return 0;}
template <typename T> istream &operator>>(istream &is, vector<T> &vec){for(auto &v: vec)is >> v; return is;}
template <typename T> ostream &operator<<(ostream &os, const vector<T>& vec){for(int i = 0; i < vec.size(); i++){ os << vec[i]; if(i + 1 != vec.size())os << " ";} return os;}
template <typename T> ostream &operator<<(ostream &os, const set<T>& st){for(auto itr = st.begin(); itr != st.end(); ++itr){ os << *itr; auto titr = itr; if(++titr != st.end())os << " ";} return os;}
template <typename T> ostream &operator<<(ostream &os, const unordered_set<T>& st){for(auto itr = st.begin(); itr != st.end(); ++itr){ os << *itr; auto titr = itr; if(++titr != st.end())os << " ";} return os;}
template <typename T> ostream &operator<<(ostream &os, const multiset<T>& st){for(auto itr = st.begin(); itr != st.end(); ++itr){ os << *itr; auto titr = itr; if(++titr != st.end())os << " ";} return os;}
template <typename T> ostream &operator<<(ostream &os, const unordered_multiset<T>& st){for(auto itr = st.begin(); itr != st.end(); ++itr){ os << *itr; auto titr = itr; if(++titr != st.end())os << " ";} return os;}
template <typename T1, typename T2> ostream &operator<<(ostream &os, const pair<T1, T2> &p){os << p.first << " " << p.second; return os;}
template <typename T1, typename T2> ostream &operator<<(ostream &os, const map<T1, T2> &mp){for(auto itr = mp.begin(); itr != mp.end(); ++itr){ os << itr->first << ":" << itr->second; auto titr = itr; if(++titr != mp.end())os << " "; } return os;}
template <typename T1, typename T2> ostream &operator<<(ostream &os, const unordered_map<T1, T2> &mp){for(auto itr = mp.begin(); itr != mp.end(); ++itr){ os << itr->first << ":" << itr->second; auto titr = itr; if(++titr != mp.end())os << " "; } return os;}
//  types
using ll = long long int;
using P = pair<int, int>;
using Pli = pair<ll, int>;
using Pil = pair<int, ll>;
using Pll = pair<ll, ll>;
using Pdd = pair<double, double>;
using cd = complex<double>;
// constants
const int inf = 1e9;
const ll linf = 1LL << 50;
const double EPS = 1e-10;
const int mod = 1000000007;
const int dx[4] = {-1, 0, 1, 0};
const int dy[4] = {0, -1, 0, 1};
// io
struct fast_io{
  fast_io(){ios_base::sync_with_stdio(false); cin.tie(0); cout << fixed << setprecision(20);}
} fast_io_;


int main(int argc, char const* argv[])
{
  int n; cin >> n;
  vector<ll> l(n); cin >> l;
  int q; cin >> q;
  vector<int> k(q); cin >> k;
  int K = 0; rep(i, q)K = max(K, k[i]);

  priority_queue<pair<double, P>> pq;
  rep(i, n)pq.push(mk(l[i], mk(i, 1)));
  vector<double> ans(K + 1, 0);
  for(int i = 1; i <= K; i++){
    auto p = pq.top(); pq.pop();
    ans[i] = p.first;
    pq.push(mk(l[p.second.first] / (double)(p.second.second + 1), mk(p.second.first, p.second.second + 1)));
  }
  rep(i, q){
    cout << ans[k[i]] << endl;
  }
  return 0;
}
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