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<iostream>
using namespace std;
#include<algorithm>
#include<vector>
#include<set>
#include<map>
#include<cmath>
#define int long long int
int binary_search(int arr[],int n,int start,int end,int k)
{
	if(start>end)
	    return -1;
	else
	{
		int mid=(start+end)/2;
		int total=0;
		int j;
		for(j=0;j<n;j++)
		{
			int here=arr[j]/mid;
			if(arr[j]%mid==0)
			   here--;
			total+=here;
		}
		if(total<=k)
		{
			int x=mid;
			int y=binary_search(arr,n,start,mid-1,k);
			if(y==-1)
			   return x;
			else
			   return y;
		}
		else
		   return binary_search(arr,n,mid+1,end,k);
	}
}
int32_t main()
{
	ios_base::sync_with_stdio(false);
	cin.tie(NULL);
	cout.tie(NULL);
	int n,k;
	cin>>n>>k;
	int j,arr[n];
	for(j=0;j<n;j++)
	   cin>>arr[j];
	int answer=binary_search(arr,n,1,10000000000,k);
	cout<<answer<<endl;
}