# Educational Codeforces Round 44 (Rated for Div. 2)A. Chess Placing

A. Chess Placing

time limit per test

1 second

memory limit per test

256 megabytes

input

standard input

output

standard output

You are given a chessboard of size 1 × n. It is guaranteed that n is even. The chessboard is painted like this: "BWBW...BW".

Some cells of the board are occupied by the chess pieces. Each cell contains no more than one chess piece. It is known that the total number of pieces equals to

.

In one step you can move one of the pieces one cell to the left or to the right. You cannot move pieces beyond the borders of the board. You also cannot move pieces to the cells that are already occupied.

Your task is to place all the pieces in the cells of the same color using the minimum number of moves (all the pieces must occupy only the black cells or only the white cells after all the moves are made).

Input

The first line of the input contains one integer n (2 ≤ n ≤ 100, n is even) — the size of the chessboard.

The second line of the input contains

integer numbers

(1 ≤ pi ≤ n) — initial positions of the pieces. It is guaranteed that all the positions are distinct.

Output

Print one integer — the minimum number of moves you have to make to place all the pieces in the cells of the same color.

Examples

input

Copy

```6
1 2 6```

output

Copy

`2`

input

Copy

```10
1 2 3 4 5```

output

Copy

`10`

Note

In the first example the only possible strategy is to move the piece at the position 6 to the position 5 and move the piece at the position 2to the position 3. Notice that if you decide to place the pieces in the white cells the minimum number of moves will be 3.

In the second example the possible strategy is to move

in 4 moves, then

in 3 moves,

in 2 moves and

in 1 move.

#include <iostream> #include <cmath> #include <algorithm> using namespace std; int main() { int n,ans1=0,ans2=0; int a[105]; scanf("%d",&n); for(int i=1;i<=n/2;i++)   scanf("%d",&a[i]);   sort(a+1,a+n/2+1); for(int i=1,j=1,k=2;i<=n/2;i++,j+=2,k+=2) { ans1+=abs(a[i]-j); ans2+=abs(a[i]-k); } printf("%d\n",min(ans1,ans2)); return 0;  }

309 篇文章26 人订阅

0 条评论

## 相关文章

### Java_数据交换_JAXB_用法入门

JAXB（Java Architecture for XML Binding) 是一个业界的标准，是一项可以根据XML Schema产生Java类的技术。该过程...

1053

1944

1891

### 深入分析Spring MVC中RequestBody与ResponseBody

在SpringMVC中，可以使用@RequestBody和@ResponseBody两个注解，分别完成请求报文到对象和对象到响应报文的转换。在Sprin...

2581

3115

1002

### VF

Vasya is the beginning mathematician. He decided to make an important contributi...

881

### HDUOJ----(1030)Delta-wave

Delta-wave Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K...

3447

1K5

1181