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 C (GCC 9.2.1) (judge verdict: Accepted)

#include <stdio.h>
#include <math.h>
#include <assert.h>
#include <stdlib.h>
#include <limits.h>

void binary_search(int n,int k,int a[n],int *min,int *max);

int main(void)
{

	int n,k;
	int *a;
	int maxa=INT_MIN;
	int mina=1;
	int x;
	
	scanf("%d %d",&n,&k);
	a=(int *)malloc(sizeof(int)*(n));
	
	int i;
	for(i=0;i<n;i++)
	{
		scanf("%d",&a[i]);
	}
	
	for(i=0;i<n;i++)
	{
		if(maxa<a[i])
		{
			maxa=a[i];
		}
	}
	
	binary_search(n,k,a,&mina,&maxa);
	printf("%d",mina);
	
	free(a);
	a=NULL;
	return 0;
}

void binary_search(int n,int k,int a[n],int *min,int *max)
{
	int mid=((*min)+(*max))/2;
	int b=0;
	int i;
	for(i=0;i<n;i++)
	{
		b+=((a[i]+mid-1)/mid)-1;
	}
	if(b<=k)
	{
		(*max)=mid;
	}
	else
	{
		(*min)=mid+1;
	}
	if((*min)!=(*max))
	{
		binary_search(n,k,a,min,max);
	}
	
}