Problem:
Score : 400 points

Problem Statement

You are given N integers; the i-th of them is A_i.
Find the maximum possible sum of the absolute differences between the adjacent elements after arranging these integers in a row in any order you like.

Constraints

2 \leq N \leq 10^5

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
A_1
:
A_N

Output

Print the maximum possible sum of the absolute differences between the adjacent elements after arranging the given integers in a row in any order you like.

Sample Input 1

5
6
8
1
2
3

Sample Output 1

21

When the integers are arranged as 3,8,1,6,2, the sum of the absolute differences between the adjacent elements is |3 - 8| + |8 - 1| + |1 - 6| + |6 - 2| = 21. This is the maximum possible sum.

Sample Input 2

6
3
1
4
1
5
9

Sample Output 2

25

Sample Input 3

3
5
5
1

Sample Output 3

8

Solution in C (GCC 5.4.1) (judge verdict: Wrong Answer)

#include<stdio.h>
#define max(xx,yy) (xx>yy) ? (xx) : (yy)
#define abs(zz) (zz>=0) ? (zz) : (-(zz))

void quicksort(int data[],int left,int right);

int main()
{
    int n,m,i,j,k,a,b,num[100010],ans1=0,ans2=0;
	int board[101][101]={0};
	scanf("%d",&n);
	for(i=0;i<n;i++)
	{
		scanf("%d",&(num[i]));
	}
	quicksort(num,0,n-1);
	//pattern1
	m=n/2;
	a=num[0];
	b=num[n-1];
	ans1=(abs(a-b));
	for(i=1;i<m;i++)
	{
		ans1+=abs((a-num[n-1-i]));
		ans1+=abs((b-num[i]));
		b=num[n-1-i];
		a=num[i];
	}
	//printf("%d\n",ans1);
	if(n%2==1)ans1+=max((abs(a-num[m])),(abs(b-num[m])));
	
	printf("%d",ans1);
}

void quicksort(int data[],int left,int right)
{
	int i,j;
	int std,work;
	if(left<right)
	{
		i=left,j=right;
		std=data[(i+j)/2];//基準
		do
		{
			while(data[i] < std)i++;
			while(data[j] > std)j--;
			if(i<=j)
			{
				if(i<j)
				{
					work=data[i];
					data[i]=data[j];
					data[j]=work;
				}
				i++; j--;
			}
		}while(i<=j);
		if(left<j)quicksort(data,left,j);
		if(i<right)quicksort(data,i,right);
	}
}