結果

問題 No.31 悪のミックスジュース
コンテスト
ユーザー fumiphys
提出日時 2019-08-19 21:48:01
言語 C++14
(gcc 13.3.0 + boost 1.87.0)
結果
AC  
実行時間 26 ms / 5,000 ms
コード長 4,195 bytes
コンパイル時間 1,694 ms
コンパイル使用メモリ 170,716 KB
実行使用メモリ 5,248 KB
最終ジャッジ日時 2024-10-05 05:29:52
合計ジャッジ時間 2,735 ms
ジャッジサーバーID
（参考情報） 		judge2 / judge1
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ファイルパターン 結果
other AC * 17
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>

// macros
#define pb emplace_back
#define mk make_pair
#define pq priority_queue
#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 vrep(v, i) for(int i = 0; i < (v).size(); i++)
#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 FI first
#define SE second
#define bit(n) (1LL<<(n))
#define INT(n) int n; cin >> n;
#define LL(n) ll n; cin >> n;
#define DOUBLE(n) double n; cin >> n;
using namespace std;

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 ll linf = numeric_limits<ll>::max();
const double EPS = 1e-10;
const int mod = 1e9 + 7;
const int dx[4] = {-1, 0, 1, 0};
const int dy[4] = {0, -1, 0, 1};

// solve
ll dp[2000010];


int main(int argc, char const* argv[])
{
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  cout << fixed << setprecision(20);
  INT(n); LL(v);
  vector<ll> c(n); cin >> c;
  vector<ll> cs(n, 0); rep(i, n)cs[i] = (i > 0 ? cs[i-1]: 0) + c[i];
  if(v <= n){
    cout << cs[n-1] << endl;
    return 0;
  }
  v -= n;

  int m = 200000;
  rep(i, m + 1)dp[i] = linf; dp[0] = 0;
  rep(i, n){
    rep(j, m + 1){
      if(j - i - 1 >= 0)dp[j] = min(dp[j], dp[j - i - 1] + cs[i]);
    }
  }
  for(int i = m - 1; i >= 0; i--){
    dp[i] = min(dp[i], dp[i+1]);
  }

  double mini = 1e18;
  int at = 0;
  rep(i, n){
    double tmp = (double(cs[i])) / (double(i + 1));
    if(mini > tmp){
      mini = tmp;
      at = i;
    }
  }
  ll res = linf;
  for(int i = 0; i <= m; i++){
    ll tmp = cs[n-1];
    ll tar = v - i;
    if(tar < 0)break;
    ll req = (tar + at) / (at + 1);
    ll rem = v - req * (at + 1);
    tmp += req * cs[at];
    if(rem > 0 && rem <= m)tmp += dp[rem];
    res = min(res, tmp);
  }
  cout << res << endl;
  return 0;
}
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