For this Question: There are n ropes, you need to cut k pieces of the same length from them. Find the maximum length of the piece you can get.The first line of the input contains two integers n and k. 
Next n lines contain one number each, the length of the rope a<sub>i</sub>.

<b>Constraints</b>
1 &le; n, k &le; 10000
1 &le; a<sub>i</sub> &le; 10<sup>7</sup>Print one real number maximum length of the piece you can get. Print the result up to 6 decimal places.Sample Input 
4 11
802
743
457
539

Sample Output
200.500000, I have written this Solution Code: #include <bits/stdc++.h>

#include <ext/pb_ds/assoc_container.hpp>
#include <ext/pb_ds/tree_policy.hpp>

using namespace std;
using namespace __gnu_pbds;

template <class T>
using ordered_set = tree<T, null_type, less<T>, rb_tree_tag, tree_order_statistics_node_update>;

template <class key, class value, class cmp = std::less<key>>
using ordered_map = tree<key, value, cmp, rb_tree_tag, tree_order_statistics_node_update>;
// find_by_order(k)  returns iterator to kth element starting from 0;
// order_of_key(k) returns count of elements strictly smaller than k;

#ifdef LOCAL
#define debug(...) cerr << "[" << #__VA_ARGS__ << "]:", debug_out(__VA_ARGS__)
#else
#define debug(...) 2351
#endif

#define int long long

mt19937_64 rng(chrono::steady_clock::now().time_since_epoch().count());
inline int64_t random_long(int l = LLONG_MIN, int r = LLONG_MAX) {
    uniform_int_distribution<int64_t> generator(l, r);
    return generator(rng);
}

int32_t main() {
    ios_base::sync_with_stdio(0);
    cin.tie(0);
    cout.tie(0);
    auto start = std::chrono::high_resolution_clock::now();

    int n, k;
    cin >> n >> k;
    vector<double> a(n);
    for (int i = 0; i < n; i++) {
        cin >> a[i];
    }
    double l = 0, r = 1e9;
    auto check = [&](double x) -> bool {
        int piece = 0;
        for (int i = 0; i < n; i++) {
            piece += a[i] / x;
        }
        if (piece >= k) {
            return true;
        }
        return false;
    };

    while (r - l > 0.000001) {
        double mid = l + (r - l) / 2;
        if (check(mid))
            l = mid;
        else
            r = mid;
    }
    printf("%0.6f", l);

    return 0;
};
, In this Programming Language: C++, Now tell me if this Code is compilable or not?