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<bits/stdc++.h>
#include<ext/pb_ds/assoc_container.hpp>
#include<ext/pb_ds/tree_policy.hpp>
#define int long long
#define mkp make_pair
#define pb push_back
#define ff first
#define ss second
#define debug1(a) cout<<a<<endl;
#define debug2(a,b) cout<<a<<' '<<b<<endl;
#define debug3(a,b,c) cout<<a<' '<<b<<' '<<c<<endl;
#define rep(i,n) for(int i=0;i<n;i++)
#define repre(i,a,b)for(int i=a;i<=b;i++)
#define clr1(arr) memset(arr,-1,sizeof(arr));
#define clr0(arr) memset(arr,0,sizeof(arr));
#define pi pair<int,int>
#define pii pair<int,pi>
#define aint(v) v.begin(),v.end()
#define fastio ios_base::sync_with_stdio(false);cin.tie(NULL);cout.tie(NULL);
using namespace __gnu_pbds;
using namespace std;
typedef tree<int,null_type,less<int>,rb_tree_tag,tree_order_statistics_node_update> orderedSet;
typedef tree<int,null_type,less_equal<int>,rb_tree_tag,tree_order_statistics_node_update> orderedMSet;
struct custom_hash {
static uint64_t splitmix64(uint64_t x) {
        // http://xorshift.di.unimi.it/splitmix64.c
        x += 0x9e3779b97f4a7c15;
        x = (x ^ (x >> 30)) * 0xbf58476d1ce4e5b9;
        x = (x ^ (x >> 27)) * 0x94d049bb133111eb;
        return x ^ (x >> 31);
    }

    size_t operator()(uint64_t x) const {
        static const uint64_t FIXED_RANDOM = chrono::steady_clock::now().time_since_epoch().count();
        return splitmix64(x + FIXED_RANDOM);
    }
};
#define ump gp_hash_table<int,int,custom_hash>
//*p.find_by_order(index) return value at index
//p.order_of_key(key) return index
/*int **dp,in main function,int row,col;dp=new int*[n];rep(i,n)dp[i]=new int[m]*/
//**** Maths Section*******/
int power(int x,int y);
bool isPrime(int n);
int modInv(int a,int b);
int gcdExtended(int a,int b,int *x,int* y);
int mpower(int a,int b,int p);
/***************************/
int n,k;
int maxe=0;
int arr[200009];
int check(int mid)
{
  int ans=0;
  repre(i,1,n)
  {
    ans+=(arr[i]-1)/mid;
  }
  return ans;
}
int32_t main()
{
//freopen("input.txt", "r", stdin);
//freopen("output.txt", "w", stdout);
  fastio
  cin>>n>>k;
  repre(i,1,n)
  {
    cin>>arr[i];
    maxe=max(maxe,arr[i]);
  }
  if(k==0)
    cout<<maxe<<endl;
  else
  {
    int low=1;
    int high=maxe;
    while(low<high)
    {
      int mid=(low+high)/2;
      if(check(mid)<=k)
        high=mid;
      else low=mid+1;
    }
    cout<<low<<endl;
  }

  return 0;
}

int modInv(int a, int m)
{
int x, y;
int g = gcdExtended(a, m, &x, &y);
int res = (x%m + m) % m;
return res;
}

int gcdExtended(int a, int b, int *x, int *y)
{
if (a == 0){
    *x = 0, *y = 1;
    return b;
  }
  int x1, y1;
  int gcd = gcdExtended(b%a, a, &x1, &y1);
  *x = y1 - (b/a) * x1;
  *y = x1;
  return gcd;
}

int mpower(int x, int y, int p)
{
int res = 1;
x = x % p;
while(y > 0){
    if(y & 1) res = (res*x) % p;
      y = y>>1;
      x = (x*x) % p;
    }
    return res;
}
int power(int x, int y)
{
int res = 1;
while (y > 0){
    if (y & 1) res = res*x;
      y = y>>1;
      x = x*x;
    }
    return res;
}

bool isPrime(int n)
{
if (n <= 1)  return false;
  if (n <= 3)  return true;
    if (n%2 == 0 || n%3 == 0) return false;
      int p=sqrt(n);
      for(int i=5;i<=p;i=i+6)
        if (n%i == 0 || n%(i+2) == 0)
          return false;
      return true;
}