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: Wrong Answer)

//use itertools::Itertools;
use std::cmp;
use std::collections::BTreeMap;
use std::collections::BTreeSet;
use std::collections::BinaryHeap;
use std::collections::HashMap;
use std::collections::HashSet;
use std::collections::VecDeque;
use std::io::Read;
use std::usize::MAX;

macro_rules! input {(source = $s:expr, $($r:tt)*) => {let mut iter = $s.split_whitespace();let mut next = || { iter.next().unwrap() };input_inner!{next, $($r)*}};($($r:tt)*) => {let stdin = std::io::stdin();let mut bytes = std::io::Read::bytes(std::io::BufReader::new(stdin.lock()));let mut next = move || -> String{bytes.by_ref().map(|r|r.unwrap() as char).skip_while(|c|c.is_whitespace()).take_while(|c|!c.is_whitespace()).collect()};input_inner!{next, $($r)*}};}
macro_rules! input_inner {($next:expr) => {};($next:expr, ) => {};($next:expr, $var:ident : $t:tt $($r:tt)*) => {let $var = read_value!($next, $t);input_inner!{$next $($r)*}};}
macro_rules! read_value {($next:expr, ( $($t:tt),* )) => {( $(read_value!($next, $t)),* )};($next:expr, [ $t:tt ; $len:expr ]) => {(0..$len).map(|_| read_value!($next, $t)).collect::<Vec<_>>()};($next:expr, chars) => {read_value!($next, String).chars().collect::<Vec<char>>()};($next:expr, usize1) => {read_value!($next, usize) - 1};($next:expr, [ $t:tt ]) => {{let len = read_value!($next, usize);(0..len).map(|_| read_value!($next, $t)).collect::<Vec<_>>()}};($next:expr, $t:ty) => {$next().parse::<$t>().expect("Parse error")};}

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

    let mut a = a;
    a.sort();
    a.reverse();

    let mut ans = 0;

    let mut min = 0.0;
    let mut max = 0;
    for i in 0..n {
        max = cmp::max(max, a[i]);
    }
    let mut max = max as f64;

    while max - min > 0.00001 {
        let mid = (min + max) / 2.0;
        if check_Cutout(&a, mid) > k {
            min = mid;
        } else {
            max = mid;
        }
    }

    println!("{}", min.ceil() as usize);
}

// k回でcut_lengthに切り分けられるか
fn check_Cutout(a: &[usize], cut_length: f64) -> usize {
    let mut cut_count = 0;
    for i in 0..a.len() {
        if a[i] as f64 > cut_length {
            cut_count += (a[i] as f64 / cut_length).ceil() as usize - 1;
        } else {
            break;
        }
    }

    //println!("{}: cut_count {}", cut_length, cut_count);

    cut_count
}

fn main() {
    let thd = std::thread::Builder::new().stack_size(104_857_600);
    thd.spawn(|| solve()).unwrap().join().unwrap();

    /*
    // 入力を一括で読み込む場合
    let mut buf = String::new();
    std::io::stdin().read_to_string(&mut buf).unwrap();
    let mut input = buf.split_whitespace();

    // inputに対しnext()で読み込んでいく
    let q: usize = input.next().unwrap().parse().unwrap();
    */
}

const LARGE_PRIME: u64 = 1_000_000_007;

// @〜でsnippet
// 二分探索 vは単調増加 v[x] >= aとなる最小のxを返す
fn bin_search(v: &[usize], a: usize) -> usize {
    if v.is_empty() || a <= v[0] {
        return 0;
    }
    let mut min = 0;
    let mut max = v.len();
    while max - min > 1 {
        let mid = (min + max) / 2;
        if v[mid] < a {
            min = mid;
        } else {
            max = mid;
        }
    }
    max
}