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 <cstdio>
#include <algorithm>

using namespace std;

int N;
typedef long long int ll;
ll K;
ll A[200007];

bool possible(ll max_val){
    // printf("testing %lld\n",max_val);
    ll c=0;
    for(int i=1;i<=N;i++){
        ll t=max_val*2+1;
        ll twice=A[i]*2;
        c+=((twice/t));
        // printf("c=%lld\n",c);
    }

    return c<=K;
}

int main(){

    scanf("%d%lld",&N,&K);
    for(int i=1;i<=N;i++) scanf("%lld",A+i);

    ll low=0,high=(*max_element(A+1,A+N+1))+1;
    ll res=high-1;
    // int i=0;
    while (high!=low){
        // printf("low=%lld high=%lld\n",low,high);
        ll mid=(low+high)/2;
        if (possible(mid))
            high=mid,res=mid;
        else 
            low=mid+1;
        // i++;
    }

    printf("%lld\n",low);

    return 0;
}