思路：
  题目中

Don’t forget that it is possible to have ai=0 after some operations, thus the answer always exists.

那么可以考虑用vector储存暴力枚举输入的数通过除2向下取整到0 ~ 2e^5的次数；
然后对每种情况从小到大排序，再取出前k个的和进行比较；

ACcode：

/*
	author:nttttt;
	add oil!
*/

#include<iostream>
#include<cstdio>
#include<cstring>
#include<queue>
#include<algorithm>
#include<stack>
#include<string>
#include<utility>
#include<cmath>
#include<vector>
#include<functional>//使用 greater<int>();
using namespace std;
typedef long long ll;
typedef pair<int,int> pll;
const int INF = 0x3f3f3f3f;
const int N = 1e6 + 10;
vector<int> a[N];
int b[N];
int main()
{
	int n,k;
	cin >> n >> k;
	for(int i = 1; i <= n ;i++)
	{
		ll cnt = 0;
		cin >> b[i];
		while(b[i])
		{
			a[b[i]].push_back(cnt++);
			b[i] /= 2;	
		} 		
	}
	
	ll t ,res = INF;
	for(int i = 0; i <= 200010;i++)
	{
		t = 0;
		if(a[i].size() < k) continue;
		sort(a[i].begin(),a[i].end());
		for(int j = 0; j < k;j++)
			t += a[i][j];
		res = min(res,t);						
	}
	cout << res << endl;	
	return 0;
}