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>
#include<vector>
#include<math.h>
#include<string.h>
using namespace std;
#define MOD 1000000007
#define MAX 200005
#define PMAX 55
#define PRECISION 0.000001
#define INF 2000000000
#define FASTIO ios_base::sync_with_stdio(false);cin.tie(NULL)
#include <ext/pb_ds/assoc_container.hpp>
#include <ext/pb_ds/tree_policy.hpp>
#include <ext/pb_ds/detail/standard_policies.hpp>
int A[MAX];
bool binsearch(double length,int n,int k)
{
    long long cuts=0;
    int i;
    double temp;
    for(i=0;i<n;i++)
    {
        temp=A[i];
        temp=temp/length;
        //temp=temp-0.00000001;
        cuts=cuts+(long long) temp;
    }
    return cuts<=k;
}
int main()
{
    int n,i,k;
    scanf("%d %d",&n,&k);
    int total=0;
    for(i=0;i<n;i++)
    {
        scanf("%d",&A[i]);
        total=max(total,A[i]);
    }
    double first=0,last=total;
    double middle=(first+last)/2;
    while(last-first>PRECISION)
    {
        //printf("first=%.10f, last=%.10f, middle=%.10f\n",first,last,middle);
        if(binsearch(middle,n,k)) last=middle;
        else first=middle;
        middle=(first+last)/2;
    }
    printf("%.0f",middle);
    return 0;
}