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<stdlib.h>
#include<string.h>
#include<math.h>
#define NUM 200020
#define MAX 1000000010

int stock[NUM]={0};

//長さlenの丸太が何本とれるか
int available(int len, int n)
{
    int j;
    int r=0;
    for(j=0;j<n;j++)
    {
        r+= ceil((double)stock[j]/(double)len);
    }
    //printf("len:%d ->%d\n",len,r);
    return r;
}

int main() {
    int n = 0,k=0;
    int i, j;
    int x = 0;
    scanf("%d %d", &n,&k);

    for (i = 0; i < n; i++) {
        scanf("%d", &(stock[i]));
    }

    //にぶたん
    int hi,lo,mid;
    hi = MAX;
    lo =0;

    while(hi-1 > lo)
    {
        mid =(hi+lo)/2;
        if(available(mid,n) > k+n){
            lo = mid;
        }
        else{
            hi = mid;
        }
    }
    if(k==0)
    {
        int max=0;
        for(i=0;i<n;i++)
        {
            if(stock[i]>max)max=stock[i];
        }
        printf("%d",max);
        return 0;
    }
    /*
    else
    {
        int l;
        for(l=lo;l>0;l--)
        {
            if(available(l,n)!=(n+k))break;
        }
        printf("%d",l+1);
        return 0;
    }
     */
    printf("%d",lo+1);
}