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<bits/stdc++.h>
using namespace std;
const int MAXN = 2e5 + 5;
int n, k;
double a[MAXN];
bool Check(double x){
	priority_queue<double> p;
	for(int i = 1; i <= n; i++) p.push(a[i]);
	for(int i = 1; i <= k; i++){
		double now = p.top();
		p.pop();
		if(now <= x) break;
		p.push(x);
		p.push(now - x);
	}
	if(p.top() > x) return false;
	else return true;
}
inline int read(void){
	int num = 0, f = 1;
	char ch;
	while(!isdigit(ch=getchar())) if(ch == '-') f = -1;
	while(isdigit(ch)) num = num*10 + ch-'0', ch = getchar();
	return num * f;
}
int main()
{
	n = read(), k = read();
	for(int i = 1; i <= n; i++) a[i] = 1.0 * read();
	double l = 1, r = 1e9, ans;
	while(r - l > 0.001){
		double mid = (l+r) / 2;
		if(Check(mid)){
			ans = mid, r = mid;
		}
		else l = mid;
	}
	printf("%.0f\n", ceil(ans));
	return 0;
}