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: Accepted)

import java.util.Scanner;

public class Main {

	public static void main(String[] args) {

		Scanner scan = new Scanner(System.in);

		//入力値取得
		int N = scan.nextInt();
		long K = scan.nextLong();

		long[] A = new long[N];

		double left = 0.0;
		double right = 0.0;

		for (int i = 0; i < N; i++) {
			A[i] = scan.nextLong();
			if (A[i] > right) {
				right = A[i];
			}
		}

		//演算
		if (K != 0L) {
			while (right - left > 0.000001) {
				double mid = (left + right) / 2.0;

				long result = 0L;

				for (int i = 0; i < N; i++) {
					result += (long) Math.ceil(A[i] / mid) - 1L;
				}

				if (result > K) {
					left = mid;
				} else {
					right = mid;
				}
			}
		}

		//結果出力
		System.out.println((long) Math.ceil(right));
	}
}