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: Time Limit Exceeded)

use proconio::{fastout, input};
use std::collections::BinaryHeap;
use ordered_float::NotNan;

#[fastout]
fn main() {
    input! {
        n: usize,
        k: i64,
        a: [i64; n],
    }

    let mut q = BinaryHeap::new();
    for a_i in a.iter().copied() {
        q.push((NotNan::new(a_i as f64).unwrap(), 1i64, a_i));
    }

    for _ in 0..k {
        let (_, c, a_i) = q.pop().unwrap();
        let c = c + 1;
        q.push((NotNan::new(a_i as f64 / c as f64).unwrap(), c, a_i));
    }

    let (ans, _, _) = q.pop().unwrap();
    println!("{}", ans.ceil());

    // let s: i64 = a.iter().sum();
    // // let avg = s / (k + n);
    // let mut ans = std::f64::INFINITY;
    // for ai in a.iter().copied() {
    //     let m = ai * (k + n as i64) / s;
    //     dbg!(m);
    //     ans = f64::min(ans, ai as f64 / m as f64);
    // }
    // println!("{}", ans.ceil());
}