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);
        int N = sc.nextInt();
        long K = sc.nextLong();
        double[] A = new double[N];
        double max = 0;
        for (int i=0;i<N;i++) {
            A[i] = sc.nextInt()+0.005;
            max = Math.max(max, A[i]);
        }

        // double ok = 10_000_000_000L;
        double ok = max;
        double ng = 0L;

        while (ok-ng > 0.00001) {
            double mid = (ok+ng)/2;
            long cnt = 0L;
            for (int i=0;i<N;i++) {
                cnt += Math.floor(A[i]/(mid+0.00000000005));
            }
            // System.out.println("mid, cnt: "+mid+" "+cnt);

            if (cnt > K) {
                ng = mid;
            } else { // cnt <= K
                ok = mid;
            }
        }
        // System.out.println(ok);
        System.out.println((int)(Math.ceil(ok-0.05)));
    }
}