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>
#define ll long long
const ll MAX=100050;
const ll modn=1e9+7;
using namespace std;
ll a[200050];
ll n,k;
ll ans;
void fz(ll l,ll r){
    //cout<<l<<"  "<<r<<endl;
    if(l>=r) {
        ans=l;
        return;
    }
    ll m=(l+r)/2;
    ll sum=0;
    for(int i=1;i<=n;i++){
        sum+=a[i]/m;
        if(a[i]%m==0)
            sum--;
    }
    if(sum<=k)
        fz(l,m);
    else fz(m+1,r);
}
void slove(){

    cin>>n>>k;
    ll maxn=0;
    for(int i=1;i<=n;i++) {
        cin >> a[i];
        maxn=max(maxn,a[i]);
    }
    fz(1,maxn);
    cout<<ans;
}

int main() {
    int _;
    //scanf("%d",&_);
    _=1;
    while(_--) {
        slove();
    }
    return 0;
}