Problem:
Score : 500 points

Problem Statement

We have N logs of lengths A_1,A_2,\cdots A_N.

We can cut these logs at most K times in total. When a log of length L is cut at a point whose distance from an end of the log is t (0<t<L), it becomes two logs of lengths t and L-t.

Find the shortest possible length of the longest log after at most K cuts, and print it after rounding up to an integer.

Constraints

1 \leq N \leq 2 \times 10^5

0 \leq K \leq 10^9

1 \leq A_i \leq 10^9

All values in input are integers.

Input

Input is given from Standard Input in the following format:

N K
A_1 A_2 \cdots A_N

Output

Print an integer representing the answer.

Sample Input 1

2 3
7 9

Sample Output 1

4

First, we will cut the log of length 7 at a point whose distance from an end of the log is 3.5, resulting in two logs of length 3.5 each.

Next, we will cut the log of length 9 at a point whose distance from an end of the log is 3, resulting in two logs of length 3 and 6.

Lastly, we will cut the log of length 6 at a point whose distance from an end of the log is 3.3, resulting in two logs of length 3.3 and 2.7.

In this case, the longest length of a log will be 3.5, which is the shortest possible result. After rounding up to an integer, the output should be 4.

Sample Input 2

3 0
3 4 5

Sample Output 2

5

Sample Input 3

10 10
158260522 877914575 602436426 24979445 861648772 623690081 433933447 476190629 262703497 211047202

Sample Output 3

292638192

Solution in C++ (GCC 9.2.1) (judge verdict: Wrong Answer)

//Author: Pakhomovee 
#include <iostream> 
#include <vector>
#include <string>
#include <algorithm>
#include <cmath>
#include <map>
#include <set>
#include <unordered_map>
#include <unordered_set>
#include <chrono>
#include <deque>
#include <iomanip>
#include <queue>

using namespace std;
using namespace std::chrono;

#define ld long double

int r(ld x)
{
	int z = (int)x;
	x -= z;
	if (x < 0.5)
	{
		return z;
	}
	return z + 1;
}

void solve()
{
	int n, k;
	cin >> n >> k;
	vector<int> a(n);
	for (int i = 0; i < n; ++i)
	{
		cin >> a[i];
	}
	int l = 0, r = 1e9;
	while (l + 1 < r)
	{
		int m = (l + r) / 2;
		int cnt = 0;
		for (int i = 0; i < n; ++i)
		{
			cnt += a[i] / m;
		}
		if (cnt <= k)
		{
			r = m;
		}
		else
		{
			l = m;
		}
	}
	cout << r << endl;
}

int main() {
	ios_base::sync_with_stdio(false);
	cin.tie(0);
	cout.tie(0);
	auto start = system_clock::now();

	solve();

	auto end = system_clock::now();
	cerr << "Time: " << duration_cast<milliseconds> (end - start).count() << "ms" << endl;
}