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 Python (3.8.2) (judge verdict: Runtime Error)


#!/usr/bin/python
# -*- coding: utf-8 -*-

# def
def int_mtx(N):
    x = []
    for _ in range(N):
        x.append(list(input()))
    return x

def int_map():
    return map(float,input().split())

def int_list():
    return list(map(int,input().split()))

def print_space(l):
    return print(" ".join([str(x) for x in l]))

# import
import numpy as np
from collections import Counter
from math import ceil

# main code
N,K = int_map()
A = int_list()

c = Counter(A)

lo = sorted(list(c.keys()),reverse=True)

now = lo.pop(0)
num = c.pop(now)
K = K-num

if K <= 0:
    print(ceil(max(now/2,lo[0])))
    exit()

l1 = [now/2]
c[now/2] = c[now/2] + 2*num

while True:
    if len(l1) > 0 and len(lo) > 0:
        if lo[0] >= l1[0]:
            now = lo.pop(0)
        else:
            now = l1.pop(0)
    elif len(lo) >0:
        now = lo.pop(0)
    else:
        now = l1.pop(0)

    num = c.pop(now)

    K = K-num
    l1.append(now/2)

    if K <= 0:
        if len(lo) > 0 and len(l1) > 0:
            print(ceil(max(max(now/2,lo[0]),l1[0])))
        elif len(lo) == 0:
            print(ceil(max(now/2,l1[0])))
        else:
            print(ceil(max(now/2,lo[0])))
        exit()

    c[now/2] = c[now/2] + 2*num