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++14 (Clang 3.8.0) (judge verdict: Wrong Answer)

#include <stdio.h>
#include <stdlib.h>
int int_sort(const void*a,const void *b){
    return *(int*)b-*(int*)a;
}

int main() {
    int a,tmp,sum1=0,sum2=0;
    int b[100000]={};
    int big[50000]={};
    int sma[50000]={};
    scanf("%d",&a);
    for (int i = 0; i < a; i++) {
        scanf("%d",&b[i]);
    }
    qsort(b,a,sizeof(b[0]),int_sort);
    if(a%2==0) {
        for (int i = 0; i < a; i++) {
            if (i<a/2) big[i] = b[i];
            else sma[-i+a-1] = b[i];
        }
        for(int i=0;i<a/2;i++){
            sum1+=big[i]*2-sma[i]*2;
        }
        sum1=sum1-big[0]+sma[a/2-1];

        if(big[a/2-1]-sma[a/2-1]<big[0]-sma[a/2-1])
            sum1+=big[0]-sma[a/2-1]-(big[a/2-1]-sma[a/2-1]);
    }
    else{
        for (int i = 0; i < a; i++) {
            if (i<a/2) big[i] = b[i];
            else sma[-i+a-1] = b[i];
        }
        for(int i=0;i<a/2;i++){
            sum1+=big[i]*2-sma[i]*2;
        }
        sum1=sum1-big[0]+sma[a/2-1];
        
        if(sma[a/2]-sma[a/2-1]>big[0]-sma[a/2])
            sum1+=sma[a/2]-sma[a/2-1];
        else sum1+=big[0]-sma[a/2];
    }

    printf("%d\n",sum1);
    return 0;
}