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: Time Limit Exceeded)

#include<stdio.h>
//大きい順にソート
int c[200002][2];
void merge(int i,int j,int s[][2]){
  if(i!=j){
    merge(i,(i+j)/2,s);
    merge((i+j)/2+1,j,s);
    int a=i,b=(i+j)/2+1;
    int d;
    for(d=i;d<=j;d++){
      if((i+j)/2<a){
        c[d][0]=s[b][0];
        c[d][1]=s[b][1];
        b++;
      }
      else if(j<b){
        c[d][0]=s[a][0];
        c[d][1]=s[a][1];
        a++;
      }
      else if((double)s[a][0]/(double)s[a][1]>(double)s[b][0]/(double)s[b][1]){
        c[d][0]=s[a][0];
        c[d][1]=s[a][1];
        a++;
      }
      else{
        c[d][0]=s[b][0];
        c[d][1]=s[b][1];
        b++;
      }
    }
    for(d=i;d<=j;d++){
      s[d][0]=c[d][0];
      s[d][1]=c[d][1];
    }
  }
}

int main(){
  int n,k,count=0;
  scanf("%d%d",&n,&k);
  int a[n+1][2],i;
  long long sum=0;
  for(i=0;i<n;i++){
    scanf("%d",&a[i][0]);
    sum+=(long long)a[i][0];
  }
  double max=(double)sum/(double)k;
  for(i=0;i<n;i++){
    int num=1;
    while(max*(double)num<a[i][0])
      num++;
    a[i][1]=num;
    count+=num-1;
  }
  for(i=count;i<k;i++){
      merge(0,n-1,a);
      a[0][1]++;
  }
  merge(0,n-1,a);
  if(a[0][0]%a[0][1]==0)
    printf("%d\n",a[0][0]/a[0][1]);
  else
    printf("%d\n",a[0][0]/a[0][1]+1);
  return 0;
}