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 Java (OpenJDK 11.0.6) (judge verdict: Wrong Answer)

import java.util.*;

public class Main {
    public static void main(String[] args) {
        Scanner sc = new Scanner(System.in);
        long n = sc.nextLong();
        long k = sc.nextLong();
        ArrayList<Long> list = new ArrayList<Long>();
        for (var i = 0; i < n; i++) {
            long a = sc.nextLong();
            list.add(a);
        }
        list.sort(Collections.reverseOrder());
        double ansMax = list.get(0);
        double ansMin = 1;
        double ansMid = (ansMax + ansMin) / 2;

        Solver s = new Solver();
        while (true){
            if (s.solve(list, k, ansMid)){
                ansMax = ansMid;
            } else{
                ansMin = ansMid;
            }
            ansMid = (ansMax + ansMin) / 2;
            if (Math.floor(ansMax) == Math.ceil(ansMin)){
                if (s.solve(list, k, ansMin + 1e10)){
                    System.out.println((long) Math.ceil(ansMin) + 1);
                } else {
                    System.out.println((long) Math.ceil(ansMin));
                }
                break;
            }
        }
    }
}

class Solver {
    boolean solve(List<Long> list, long k, double length){
        long count = 0;
        if (length < 1.){
            for (var a : list) {
                count += a + 1;
                if (count > k) return false;
            }
            return true;
        }
        for (var a : list){
            count += Math.max((long) Math.ceil(a / length) - 1, 0);
            if (count > k) return false;
        }
        return true;
    }
}