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: Wrong Answer)

#include <bits/stdc++.h>                
using namespace std;

#include <ext/pb_ds/assoc_container.hpp> 
#include <ext/pb_ds/tree_policy.hpp> 
using namespace __gnu_pbds; 
  
#define ordered_set tree<int, null_type,less<int>, rb_tree_tag,tree_order_statistics_node_update> 

int main() {
    ios_base::sync_with_stdio(false);
    cin.tie(NULL);
    cout.tie(NULL);
    int n;
    long long int k;
    cin>>n>>k;
    long long int cnt=0;
    long long int a[n];
    for (int i=0;i<n;i++){
        cin>>a[i];
        cnt+=a[i];
    }
    if(cnt<=k) cout<<0<<"\n";
    else{
    long long int l=1;
    long long int r=1e9;
    while (l<r-1){
        long long int m=l+(r-l)/2;
        long long int cnt=0;
        for (int i=0;i<n;i++){
            if(a[i]>m){
                cnt+=(a[i]-1)/m;
            }
        }    
        if(cnt<=k) r=m;
        else l=m+1;
    }
    if(l==0) l++;
    long long int cnt=0;
    for (int i=0;i<n;i++){
        if(a[i]>l){
            cnt+=(a[i]-1)/l;
        }
    }
    if(cnt<=k) cout<<l<<"\n";
    else cout<<l+1<<"\n";
    }
	return 0;
}