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<cmath>
using namespace std;
const int N = 1e6 + 5;
double a[N];
int n, k;
double round(double r)
{
    return floor(r + 0.5);
}
bool Check(double x) {
    int cnt = 0;
    for(int i = 1; i <= n; ++i) {
        cnt += a[i]/x;
        if(floor(a[i]/x)*x == a[i]) --cnt;    
        if(cnt > k) return false;
    }
    return true;
}
int main() {
    scanf("%d%d",&n,&k);
    double Max = -1, Min = 0x7fffffffffff;
    for(int i = 1; i <= n; ++i) { 
        scanf("%lf",&a[i]);
        if(a[i] > Max) Max = a[i];
        if(a[i] < Min) Min = a[i];
    }
    double L = 0, R = Max, Mid;
    for(int i = 1; i <= 800; ++i) {
        Mid = (L+R)/2;
        if(Check(Mid)) R = Mid;
        else L = Mid;
    }
    //printf("%f\n",Mid);
    printf("%d\n",int(round(Mid)));
return 0;
}