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 Rust (1.42.0) (judge verdict: Accepted)

#![allow(
  non_snake_case,
  unused_variables,
  unused_assignments,
  unused_mut,
  unused_imports,
  dead_code
)]
//use proconio::fastout;
use proconio::input;
use proconio::marker::*;
use std::cmp::*;
use std::collections::*;
//use std::collections::VecDeque;

fn main() {
  input! {
     N:usize,K:i64,
     mut A:[i64;N]
  }
  A.sort();
  A.reverse();
  // 一番長い丸太A[0]を c回切って、c+1 等分する。
  // A[0]/(c+1) の長さの切り上げ値を d とする。
  // 残りの切断回数で、すべてをd以下にできるかチェック。
  // 可能ならば増やし、不可能ならばcut[i]を減らす。
  if K == 0 {
    println!("{}", A[0]);
    return;
  } else if K == 1 {
    let b = (A[0] + 1) / 2;
    println!("{}", max(b, A[1]));
    return;
  }
  let mut dmax = A[0];
  let mut dmin = 0;
  while dmin + 1 < dmax {
    //println!("min={},max={}", dmin, dmax);
    let d = (dmin + dmax) / 2;
    let mut cuts = 0;
    for i in 0..N {
      cuts += ((A[i] + d - 1) / d) - 1;
    }
    if cuts <= K {
      dmax = d;
    } else {
      dmin = d;
    }
  }
  println!("{}", dmax);
}
// c=1, d=10/2=5,cuts = 1 + 1 .. OK,
// c=2, d=11/3=3,  cuts=1