ferinの競プロ帳

競プロについてのメモ

SRM676 div1 easy WaterTank

考えたこと

  • 最大の最小化なのでにぶたんをする
  • Rを決め打つとO(n)で可能かどうかの判定ができそう
  • 出すと通る

THE典型みたいなにぶたん

#include <bits/stdc++.h>

using namespace std;
using ll = long long;
using VI = vector<int>;
using VVI = vector<VI>;
using PII = pair<int, int>;

#define FOR(i, a, n) for (ll i = (ll)a; i < (ll)n; ++i)
#define REP(i, n) FOR(i, 0, n)
#define ALL(x) x.begin(), x.end()
#define PB push_back

const ll LLINF = (1LL<<60);
const int INF = (1LL<<30);
const int MOD = 1000000007;

template <typename T> T &chmin(T &a, const T &b) { return a = min(a, b); }
template <typename T> T &chmax(T &a, const T &b) { return a = max(a, b); }
template <typename T> bool IN(T a, T b, T x) { return a<=x&&x<b; }
template<typename T> T ceil(T a, T b) { return a/b + !!(a%b); }
template<class S,class T>
ostream &operator <<(ostream& out,const pair<S,T>& a){
  out<<'('<<a.first<<','<<a.second<<')';
  return out;
}
template<class T>
ostream &operator <<(ostream& out,const vector<T>& a){
  out<<'[';
  REP(i, a.size()) {out<<a[i];if(i!=a.size()-1)out<<',';}
  out<<']';
  return out;
}

int dx[] = {0, 1, 0, -1}, dy[] = {1, 0, -1, 0};

// ll�Ő錾���� �I�I�I�I�I

class WaterTank {
   public:
   double minOutputRate(vector <int> t, vector <int> x, int C)
  {
    int n = t.size();
    auto check = [&](double mid) -> bool {
      double now = 0;
      REP(i, n) {
        now += (double)x[i]*t[i];
        now -= mid*t[i];
        if(now < 0) now = 0;
        if(now > C) return false;
      }
      return true;
    };

    double lb = 0, ub = 1e6+100;
    REP(_, 100) {
      double mid = (lb+ub)/2;
      if(check(mid)) {
        ub = mid;
      } else {
        lb = mid;
      }
    }

    return lb;
  }
};