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 std::collections::BinaryHeap;
#[derive(Default)]
//NOTE
//declare variables to reduce the number of parameters for dp and dfs etc.
struct Solver {}

impl Solver {
    fn solve(&mut self) {
        let stdin = std::io::stdin();
        let mut scn = Scanner {
            stdin: stdin.lock(),
        };
        let n: usize = scn.read();
        let k: usize = scn.read();
        let a: Vec<usize> = scn.vec(n);
        let mut left = 0;
        let mut right = 1usize << 50;

        while right - left > 1 {
            let med = (right + left) / 2;
            let mut heap = BinaryHeap::from(a.clone());
            let mut m = k;
            let mut ok = true;
            while m > 0 {

                if let Some(x) = heap.pop() {
                    if x > med {
                        if x == 1 {
                            heap.push(1);
                            heap.push(1);
                            break;
                        } else {
                            let y = if x % med == 0 {
                                x / med - 1
                            } else {
                                x / med
                            };
                            if m >= y {
                                m -= y;
                            } else {
                                ok = false;
                                break;
                            }
                        }
                    } else {
                        heap.push(x);
                        break;
                    }
                }
            }
            //eprintln!("{}, {:?}", med, heap);
            if let Some(p) = heap.pop() {
                if p > med {
                    ok = false;
                }
            }
            if ok {
                right = med;
            } else {
                left = med;
            }
        }
        println!("{}", right);
    }
}

fn main() {
    std::thread::Builder::new()
        .stack_size(64 * 1024 * 1024) // 64MB
        .spawn(|| {
            let mut solver = Solver::default();
            solver.solve()
        })
        .unwrap()
        .join()
        .unwrap();
}
pub mod utils {
    const DYX: [(isize, isize); 8] = [
        (0, 1),   //right
        (1, 0),   //down
        (0, -1),  //left
        (-1, 0),  //top
        (1, 1),   //down right
        (-1, 1),  //top right
        (1, -1),  //down left
        (-1, -1), //top left
    ];

    pub fn try_adj(y: usize, x: usize, dir: usize, h: usize, w: usize) -> Option<(usize, usize)> {
        let ny = y as isize + DYX[dir].0;
        let nx = x as isize + DYX[dir].1;
        if ny >= 0 && nx >= 0 {
            let ny = ny as usize;
            let nx = nx as usize;
            if ny < h && nx < w {
                Some((ny, nx))
            } else {
                None
            }
        } else {
            None
        }
    }
}

//snippet from kenkoooo
pub struct Scanner<R> {
    stdin: R,
}

impl<R: std::io::Read> Scanner<R> {
    pub fn read<T: std::str::FromStr>(&mut self) -> T {
        use std::io::Read;
        let buf = self
            .stdin
            .by_ref()
            .bytes()
            .map(|b| b.unwrap())
            .skip_while(|&b| b == b' ' || b == b'\n' || b == b'\r')
            .take_while(|&b| b != b' ' && b != b'\n' && b != b'\r')
            .collect::<Vec<_>>();
        unsafe { std::str::from_utf8_unchecked(&buf) }
            .parse()
            .ok()
            .expect("Parse error.")
    }
    pub fn read_line(&mut self) -> String {
        use std::io::Read;
        let buf = self
            .stdin
            .by_ref()
            .bytes()
            .map(|b| b.unwrap())
            .skip_while(|&b| b == b'\n' || b == b'\r')
            .take_while(|&b| b != b'\n' && b != b'\r')
            .collect::<Vec<_>>();
        unsafe { std::str::from_utf8_unchecked(&buf) }
            .parse()
            .ok()
            .expect("Parse error.")
    }
    pub fn vec<T: std::str::FromStr>(&mut self, n: usize) -> Vec<T> {
        (0..n).map(|_| self.read()).collect()
    }

    pub fn chars(&mut self) -> Vec<char> {
        self.read::<String>().chars().collect()
    }
}