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: Runtime Error)

import java.util.Scanner;

public class Main {

    static int N;
    static long K;
    static long[] A;

    public static void main(String[] args) {

        Scanner sc = new Scanner(System.in);
        N = sc.nextInt();
        K = sc.nextLong();
        A = new long[N + 1];
      long max2 = 0;

        for(int i = 0 ; i < N ; i++) {
        	A[i] = sc.nextLong();
        	max2 = Math.max(A[i],max2);
        }

        long max = max2;
        long min = 1l;
        if(K == 0) {
        	System.out.println(max);
        	return;
        }

        // 二分探索
        for (;;) {
            // mid
            long mid = (max + min) / 2;
            if (canSolve(mid)) {
                min = mid;
            } else {
                max = mid;
            }

            if (max - min <= 1) {
                // 終わり
                break;
            }
        }
        for(int i = (int) (min - 100) ; i < min + 100 ; i++)

        if(canSolve2(i)) {
        	System.out.println(i);
          break;
        }

    }

    // 解けるかどうか
    static boolean canSolve(long num) {
    	int sum = 0;
    	for(int i = 0 ; i < N ; i++) {
    		sum += A[i] / num;
    	}
    	return sum > K ? true : false;

    }
    static boolean canSolve2(long num) {
    	int sum = 0;
    	for(int i = 0 ; i < N ; i++) {
    		sum += A[i] / num;
    	}
    	return sum == K ? true : false;

    }
}