Problem: Example

Input

100 1 0
50 100


Output

14.57738

Solution (in CPP):

Solution:
#include <bits/stdc++.h>

using namespace std;

#define REP(i, n) for (int i=0; i<(n); ++i)
#define RREP(i, n) for (int i=(int)(n)-1; i>=0; --i)
#define FOR(i, a, n) for (int i=(a); i<(n); ++i)
#define RFOR(i, a, n) for (int i=(int)(n)-1; i>=(a); --i)

#define SZ(x) ((int)(x).size())
#define ALL(x) (x).begin(),(x).end()

#define DUMP(x) cerr<<#x<<" = "<<(x)<<endl
#define DEBUG(x) cerr<<#x<<" = "<<(x)<<" (L"<<__LINE__<<")"<<endl;

template<class T>
ostream &operator<<(ostream &os, const vector <T> &v) {
    os << "[";
    REP(i, SZ(v)) {
        if (i) os << ", ";
        os << v[i];
    }
    return os << "]";
}

template<class T, class U>
ostream &operator<<(ostream &os, const pair <T, U> &p) {
    return os << "(" << p.first << " " << p.second << ")";
}

template<class T>
bool chmax(T &a, const T &b) {
    if (a < b) {
        a = b;
        return true;
    }
    return false;
}

template<class T>
bool chmin(T &a, const T &b) {
    if (b < a) {
        a = b;
        return true;
    }
    return false;
}

using ll = long long;
using ull = unsigned long long;
using ld = long double;
using P = pair<int, int>;
using vi = vector<int>;
using vll = vector<ll>;
using vvi = vector<vi>;
using vvll = vector<vll>;

const ll MOD = 1e9 + 7;
const int INF = INT_MAX / 2;
const ll LINF = LLONG_MAX / 2;
const ld eps = 1e-9;

template<typename T>
struct edge {
    int src, to;
    T cost;
};

template<typename T>
using Graph = vector<vector<edge<T>>>;

using State = string::const_iterator;

int main() {
    cin.tie(0);
    ios::sync_with_stdio(false);
    cout << fixed << setprecision(10);

    // ifstream in("in.txt");
    // cin.rdbuf(in.rdbuf());

    double D; int N, B; cin >> D >> N >> B;
    vector<double> p(N), h(N);
    REP(i, N) cin >> p[i] >> h[i];

    double ans = numeric_limits<double>::max();
    for (int b = 1; b <= B+1; ++b) {
        double l = D / (2 * b);

        double ma = l;
        REP(i, N) {
            double x = p[i] - (ll)(p[i] / (2*l)) * (2*l);
            double t = h[i] / (-x*x / (2*l*l) + x / l);
            chmax(ma, t);
        }
        chmin(ans, sqrt(l*l / ma + ma));
    }
    cout << ans << endl;

    return 0;
}