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.*;
import java.io.*;
public class Main{
        public static void main(String[] args){
                Scanner sc = new Scanner(System.in);
                Map hs = new HashMap();
                PrintWriter ou = new PrintWriter(System.out);
                int n = Integer.parseInt(sc.next());
                int k = Integer.parseInt(sc.next());
                int[] a = new int[n];
                for(int i = 0 ; i < n ; i++){
                        a[i] = Integer.parseInt(sc.next());
                }
                sc.close();
                double le = 0;
                double ri = Math.pow(10 , 9);
                while(ri - le > 0.00001){
                        double now = (le + ri) / 2;
                        int yono = 0;
                        for(int i = 0 ; i < n ; i++){
                                yono += (int)(a[i] / now);
                        }
                        if(yono > k) le = now;
                        else ri = now;
                }
                if(ri % 1 < 0.0001 || ri % 1 > 0.999) ou.print((int)Math.round(ri) + "\n");
                else ou.print((int)Math.ceil(ri)+ "\n");
                ou.flush();
        }
}