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; } };