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 Java8 (OpenJDK 1.8.0) (judge verdict: Wrong Answer)


import java.util.Arrays;

//intのオーバーフローに気をつけろ,AGCはだいたいlongにしておけ
//毎回余りを求めることでオーバーフローしなくなる
//nextLineよりnextだ
//charは''で囲うんだぞ
//読み込みは先に全部やっておけ
//for文を回す回数に気をつけろ
	//for文は10^3の三重はTLE,10^5の二重はTLE,10^9のforはTLE
//貪欲は使えるか？？試した？？

import java.util.Scanner;

public class Main {

	public static void main(String args[]) throws InterruptedException {
		    Scanner sc = new Scanner(System.in);

		    int n = sc.nextInt();
		    int[] a = new int[n];
		    for(int i = 0; i < n; i++) {
		    	a[i] = sc.nextInt();
		    }
		    Arrays.sort(a);
		    long sa = 0;
		    for(int i = 0; i < n; i++) {
		    	if(i > 0) {
		    		if(i % 2 == 1) {
		    			sa += a[n-1-(i/2)]-a[i/2]; 
		    		}else {
		    			sa += a[n-i/2] -a[i/2];
		    		}
		    	}
		    }
		    
		    long sb = 0;
		    for(int i = 0; i < n; i++) {
		    	if(i > 0) {
		    		if(i % 2 == 1) {
		    			sb += a[n-1-i/2] - a[i/2];
		    		}else {
		    			sb += a[n-1-i/2]-a[i/2-1];
		    		}
		    	}
		    }
		    
		    System.out.println(Math.max(sa, sb));
		    
		    sc.close(); }

	public static void saiki(int a, String b) {
		//再帰コーディングのお手本

		if(a == 0) {
			System.out.println(b);
		}else {
			for(char ch = 'a'; ch <= 'c'; ch++) {
				saiki(a-1,b + ch);
			}
		}
	}

	public static  int gcd(int a, int b) {
	    if (a < b) {
	        int tmp = a;
	        a = b;
	        b = tmp;
	    }

	    int r = -1;
	    while (r != 0) {
	        r = a % b;
	        a = b;
	        b = r;
	    }

	    return a;
	}
}