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 Kotlin (1.3.71) (judge verdict: Accepted)

import kotlin.math.abs
import kotlin.math.floor
import kotlin.math.round
import kotlin.math.roundToInt

private class E {
    fun solve() {
        val n = nextInt()
        val k = nextInt()
        val a = Array(n) { nextInt() }

        var l = 1
        var r = a.max()!!
        while (r > l) {
            val m = (r + l) / 2
            var count = 0
            for (x in a) {
                if (x % m == 0) {
                    count += x / m - 1
                } else {
                    count += x / m
                }
            }
            if (count > k) {
                l = m + 1
            } else {
                r = m
            }
        }
        println(r)
    }

    var tokenBuffer = listOf<String>()
    var bufferPosition = 0

    private fun checkBuffer() {
        if (bufferPosition == tokenBuffer.size) {
            tokenBuffer = nextLine().split(" ")
            bufferPosition = 0
        }
    }

    fun nextLine() = readLine()!!
    fun nextInt(): Int {
        checkBuffer()
        return tokenBuffer[bufferPosition++].toInt()
    }

    fun nextLong(): Long {
        checkBuffer()
        return tokenBuffer[bufferPosition++].toLong()
    }

    fun nextDouble(): Double {
        checkBuffer()
        return tokenBuffer[bufferPosition++].toDouble()
    }

    fun nextToken(): String {
        checkBuffer()
        return tokenBuffer[bufferPosition++]
    }
}

fun main() {
    val reader = E().solve()

}