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: Runtime Error)

#[allow(unused_imports)]
use itertools::Itertools;
#[allow(unused_imports)]
use num::*;
use proconio::input;
#[allow(unused_imports)]
use proconio::marker::*;
#[allow(unused_imports)]
use std::collections::*;

pub fn bsearch<F>(ok: i64, ng: i64, pred: F) -> Option<i64>
where
    F: Fn(i64) -> bool,
{
    let orig_ok = ok;
    let mut ok = ok;
    let mut ng = ng;
    while (ok - ng).abs() > 1 {
        let mid = (ok + ng) / 2;
        if pred(mid) {
            ok = mid;
        } else {
            ng = mid;
        }
    }
    if ok == orig_ok {
        None
    } else {
        Some(ok)
    }
}

use ordered_float::OrderedFloat;

fn solve() {
    input! {
        n: usize, k: usize,
        av: [f64; n]
    };

    let heap = av
        .into_iter()
        .map(|a| OrderedFloat::from(a))
        .collect::<BinaryHeap<_>>();
    let res = bsearch(10.pow(9) + 1, 0, |x| {
        let max = x as f64 + 0.4;
        let mut heap = heap.clone();
        for _ in 0..k {
            let a = heap.pop().unwrap();
            let a = a.into_inner();
            if a <= max {
                break;
            }
            let b1 = a - max;
            let b2 = a - b1;
            heap.push(OrderedFloat::from(b1));
            heap.push(OrderedFloat::from(b2));
        }
        let a = heap.pop().unwrap().into_inner();
        a <= max
    });
    let res = res.unwrap();
    println!("{}", res);
}

fn main() {
    std::thread::Builder::new()
        .name("big stack size".into())
        .stack_size(256 * 1024 * 1024)
        .spawn(|| {
            solve();
        })
        .unwrap()
        .join()
        .unwrap();
}