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>
#include<stdlib.h>

int int_sort( const void * a , const void * b ) {
  /* 引数はvoid*型と規定されているのでint型にcastする */
  if( *( int * )a < *( int * )b ) {
    return -1;
  }
  else
  if( *( int * )a == *( int * )b ) {
    return 0;
  }
  return 1;
}
int main(){
	int n,i;
	long a[100000],r,l,ans;
	scanf("%d",&n);
	for(i=0;i<n;i++)scanf("%ld",&a[i]);
	qsort(a,n,sizeof(long),int_sort);
	if(n%2==1){
		//from shortest
		long tmp=0;
		int cr=0,cl=0;
		r=a[cr],l=a[cr];
		cr++;
		int count=0;
		while(cr+cl<n){
		if(count%2==0){
			if(abs(r-a[n-2-cl])+abs(l-a[n-1-cl])>
			abs(l-a[n-2-cl])+abs(r-a[n-1-cl])){
			tmp+=abs(r-a[n-2-cl])+abs(l-a[n-1-cl]);
			r=a[n-2-cl],l=a[n-1-cl];
			}
			else{
			tmp+=abs(l-a[n-1-cl])+abs(r-a[n-2-cl]);
			r=a[n-1-cl],l=a[n-2-cl];
			}
			cl+=2;
		}else{
			if(abs(r-a[cr])+abs(l-a[cr+1])>
			abs(l-a[cr+1])+abs(r-a[cr])){
			tmp+=abs(r-a[cr])+abs(l-a[cr+1]);
			r=a[cr],l=a[cr+1];
			}
			else{
			tmp+=abs(l-a[cr+1])+abs(r-a[cr]);
			r=a[cr+1],l=a[cr];
			}
			cr+=2;
		}
		}
		//then max
		long tmp2=0;
		cr=0,cl=0;
		r=a[cr],l=a[cr];
		cr++;
		count=0;
		while(cr+cl<n){
		if(count%2==1){
			if(abs(r-a[n-2-cl])+abs(l-a[n-1-cl])>
			abs(l-a[n-2-cl])+abs(r-a[n-1-cl])){
			tmp2+=abs(r-a[n-2-cl])+abs(l-a[n-1-cl]);
			r=a[n-2-cl],l=a[n-1-cl];
			}
			else{
			tmp2+=abs(l-a[n-1-cl])+abs(r-a[n-2-cl]);
			r=a[n-1-cl],l=a[n-2-cl];
			}
			cl+=2;
		}else{
			if(abs(r-a[cr])+abs(l-a[cr+1])>
			abs(l-a[cr+1])+abs(r-a[cr])){
			tmp2+=abs(r-a[cr])+abs(l-a[cr+1]);
			r=a[cr],l=a[cr+1];
			}
			else{
			tmp2+=abs(l-a[cr+1])+abs(r-a[cr]);
			r=a[cr+1],l=a[cr];
			}
			cr+=2;
		}
		}
		if(tmp>tmp2)ans=tmp;
		else ans=tmp2;
	}else{
		//from shortest
		long tmp=0;
		int cr=0,cl=0;
		r=a[cr],l=a[n-1-cl];
		tmp+=abs(r-l);
		cr++;cl++;
		int count=0;
		while(cr+cl<n){
//		if(count%2==0){
			if(abs(r-a[cr])+abs(l-a[n-1-cl])>
			abs(r-a[n-1-cl])+abs(l-a[cr])){
			tmp+=abs(r-a[cr])+abs(l-a[n-1-cl]);
			r=a[cr],l=a[n-1-cl];
			}
			else{
			tmp+=abs(r-a[n-1-cl])+abs(l-a[cr]);
			r=a[n-1-cl],l=a[cr];
			}
			cl++;cr++;
/*		}else{
			if(abs(r-a[cr])+abs(l-a[cr+1])>
			abs(l-a[cr+1])+abs(r-a[cr])){
			tmp+=abs(r-a[cr])+abs(l-a[cr+1]);
			r=a[cr],l=a[cr+1];
			}
			else{
			tmp+=abs(l-a[cr+1])+abs(r-a[cr]);
			r=a[cr+1],l=a[cr];
			}
			cr+=2;
		}
		}
		//then max
		long tmp2=0;
		cr=0,cl=0;
		r=a[cr],l=a[cr];
		cr++;
		count=0;
		while(cr+cl<n){
		if(count%2==1){
			if(abs(r-a[n-2-cl])+abs(l-a[n-1-cl])>
			abs(l-a[n-2-cl])+abs(r-a[n-1-cl])){
			tmp2+=abs(r-a[n-2-cl])+abs(l-a[n-1-cl]);
			r=a[n-2-cl],l=a[n-1-cl];
			}
			else{
			tmp2+=abs(l-a[n-1-cl])+abs(r-a[n-2-cl]);
			r=a[n-1-cl],l=a[n-2-cl];
			}
			cl+=2;
		}else{
			if(abs(r-a[cr])+abs(l-a[cr+1])>
			abs(l-a[cr+1])+abs(r-a[cr])){
			tmp2+=abs(r-a[cr])+abs(l-a[cr+1]);
			r=a[cr],l=a[cr+1];
			}
			else{
			tmp2+=abs(l-a[cr+1])+abs(r-a[cr]);
			r=a[cr+1],l=a[cr];
			}
			cr+=2;
		}*/
		}/*
		if(tmp>tmp2)ans=tmp;
		else */ans=tmp;

	}
	printf("%ld",ans);
	return 0;
}