cf472B Design Tutorial: Learn from Life

 

B. Design Tutorial: Learn from Life

time limit per test

1 second

memory limit per test

256 megabytes

input

standard input

output

standard output

One way to create a task is to learn from life. You can choose some experience in real life, formalize it and then you will get a new task.

Let's think about a scene in real life: there are lots of people waiting in front of the elevator, each person wants to go to a certain floor. We can formalize it in the following way. We have n people standing on the first floor, the i-th person wants to go to the fi-th floor. Unfortunately, there is only one elevator and its capacity equal to k (that is at most k people can use it simultaneously). Initially the elevator is located on the first floor. The elevator needs |a - b| seconds to move from the a-th floor to the b-th floor (we don't count the time the people need to get on and off the elevator).

What is the minimal number of seconds that is needed to transport all the people to the corresponding floors and then return the elevator to the first floor?

Input

The first line contains two integers n and k (1 ≤ n, k ≤ 2000) — the number of people and the maximal capacity of the elevator.

The next line contains n integers: f1, f2, ..., fn (2 ≤ fi ≤ 2000), where fi denotes the target floor of the i-th person.

Output

Output a single integer — the minimal time needed to achieve the goal.

Sample test(s)

input

3 2
2 3 4

output

8

input

4 2
50 100 50 100

output

296

input

10 3
2 2 2 2 2 2 2 2 2 2

output

8

Note

In first sample, an optimal solution is:

1. The elevator takes up person #1 and person #2.
2. It goes to the 2nd floor.
3. Both people go out of the elevator.
4. The elevator goes back to the 1st floor.
5. Then the elevator takes up person #3.
6. And it goes to the 2nd floor.
7. It picks up person #2.
8. Then it goes to the 3rd floor.
9. Person #2 goes out.
10. Then it goes to the 4th floor, where person #3 goes out.
11. The elevator goes back to the 1st floor.

 

 

n个人坐电梯……第i个人要去第a[i]层。电梯只能同时载m个人，求电梯最少要移动几层

贪心……排序完显然电梯必须要到层数最大的那一层一次，还要下来。那就在这一次取层数尽可能大的m个人。

具体做法就是排序完如果n%m!=0就不断加个层数最小的，然后统计所有i是m的倍数的a[i]

还是比较好yy出来的……但是我比较逗忘记从n层到1层只要n-1层……然后样例1过不去傻傻的调了好久

#include<cstdio>
#include<iostream>
#include<cstring>
#include<cstdlib>
#include<algorithm>
#include<cmath>
#include<queue>
#include<deque>
#include<set>
#include<map>
#include<ctime>
#define LL long long
#define inf 0x7ffffff
#define pa pair<int,int>	
using namespace std;
inline LL read()
{
    LL x=0,f=1;char ch=getchar();
    while(ch<'0'||ch>'9'){if(ch=='-')f=-1;ch=getchar();}
    while(ch>='0'&&ch<='9'){x=x*10+ch-'0';ch=getchar();}
    return x*f;
}
LL n,m,sum;
LL a[100000];
int main()
{
	n=read();m=read();
	for (int i=1;i<=n;i++)a[i]=read()-1;
	sort(a+1,a+n+1);
	while(n%m!=0)
	{
		a[++n]=a[1];
	}
	sort(a+1,a+n+1);
	for (int i=m;i<=n;i+=m)
	  sum+=2*a[i];
	printf("%lld\n",sum);
}