Submission #1303040

#	Time	Username	Problem	Language	Result	Execution time	Memory
1303040		ionut27	Mobile (BOI12_mobile)	C++20	100 / 100	452 ms	15944 KiB

mobile

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

#pragma GCC optimize("Ofast")
#pragma GCC target("avx,avx2,fma")
#pragma GCC optimization("unroll-loops")

// ============ Macros starts here ============
int recur_depth = 0;
#ifdef DEBUG
#define dbg(x) {++recur_depth; auto x_=x; --recur_depth; cerr<<string(recur_depth, '\t')<<"\e[91m"<<__func__<<":"<<__LINE__<<"\t"<<#x<<" = "<<x_<<"\e[39m"<<endl;}
#else
#define dbg(x)
#endif // DEBUG
template<typename Ostream, typename Cont>
typename enable_if<is_same<Ostream, ostream>::value, Ostream&>::type operator<<(Ostream& os, const Cont& v) {
    os << "{";
    for (auto& x : v) { os << x << ", "; }
    return os << "}";
}
template<typename Ostream, typename ...Ts>
Ostream& operator<<(Ostream& os, const pair<Ts...>& p) {
    return os << "{" << p.first << ", " << p.second << "}";
}

#define readFast                      \
    ios_base::sync_with_stdio(false); \
    cin.tie(0);                       \
    cout.tie(0);
#ifdef LOCAL
#define read() ifstream fin("date.in.txt")
#else
#define read() readFast
#endif // LOCAL
// ============ Macros ends here ============

#define fin cin
#define ll long long
#define sz(x) (int)(x).size()
#define all(v) v.begin(), v.end()
#define output(x) (((int)(x) && cout << "YES\n") || cout << "NO\n")
#define LSB(x) (x & (-x))
#define test cout << "WORKS\n";

vector<pair<double, double>> a;
int n;
double L;

bool check(double d) {
    double maxRight = 0;

    for (int i = 0; i < n; ++i) {
        double c1 = sqrt(d * d - a[i].second * a[i].second);
        double left = a[i].first - c1;
        // cout << maxRight - left << " " << -(1e-3) << '\n';

        if (maxRight >= left) {
            maxRight = max(maxRight, a[i].first + c1);
        }
    }
    return maxRight >= L;
}

int main() {
    read();
    fin >> n >> L;

    a.resize(n);
    for (int i = 0; i < n; ++i) {
        fin >> a[i].first >> a[i].second;
    }

    double l = 1, r = 1.5e9, mid;
    while (r - l > 1e-3) {
        mid = (l + r) / 2;
        if (check(mid)) {
            r = mid;
        } else {
            l = mid;
        }
    }
    cout << fixed << setprecision(4) << r << '\n';

    return 0;
} /*stuff you should look for !!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!
   * test the solution with the given example
   * int overflow, array bounds, matrix bounds
   * special cases (n=1?)
   * do smth instead of nothing and stay organized
   * WRITE STUFF DOWN
   * DON'T GET STUCK ON ONE APPROACH
~Benq~*/

• Subtask 1: 4 / 4
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• Subtask 21: 5 / 5