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<iostream>
#include<vector>
#include<algorithm>
#include<cmath>
#include<ctime>
#include<queue>

using namespace std;
using ll = long long;
using vi = vector<int>;
using vvi = vector<vector<int> >;
using pii = pair<int, int>;
using ppi = pair<pii, int>;

int f_dir[2][4] = {{-1, 0, 1, 0}, {0, 1, 0, -1}}; //↑→↓←
int e_dir[2][8] = {{-1, -1, -1, 0, 0, 1, 1, 1}, {-1, 0, 1, -1, 1, -1, 0, 1}};
bool is_in_field(int y, int x, int h, int w){
  return (y >= 0 && y < h && x >= 0 && x < w);
}

int main(){
  cin.tie(0);
  ios::sync_with_stdio(false);
  int ti = clock();
  // start-----------------------------------------------
  ll n; cin >> n;
  ll k; cin >> k;
  vector<ll> a(n);
  for(int i = 0; i < n; i++) cin >> a[i];
  ll ok = *max_element(a.begin(), a.end());
  ll ng = 0;
  ll now = 0;
  auto solve = [=](ll now){
    ll ret = 0;
    for(int i = 0; i < n; i++){
      ret += (ll)(a[i]/now);
    }
    return ret;
  };
  int t = 0;
  while(ok - ng > 1){
    t++;
    now = (ng + ok) / 2;
    if(solve(now) <= k){
      ok = now;
    }
    else{
      ng = now;
    }
  }
  cout << (ll)ceil(ok) << endl;

  // end-----------------------------------------------
  // cerr << 1.0 * (clock() - ti) / CLOCKS_PER_SEC << endl;
}