Problem:
Score : 200 points

Problem Statement

Evi has N integers a_1,a_2,..,a_N. His objective is to have N equal integers by transforming some of them.

He may transform each integer at most once. Transforming an integer x into another integer y costs him (x-y)^2 dollars. Even if a_i=a_j (i≠j), he has to pay the cost separately for transforming each of them (See Sample 2).

Find the minimum total cost to achieve his objective.

Constraints

1≦N≦100

-100≦a_i≦100

Input

The input is given from Standard Input in the following format:

N
a_1 a_2 ... a_N

Output

Print the minimum total cost to achieve Evi's objective.

Sample Input 1

2
4 8

Sample Output 1

8

Transforming the both into 6s will cost (4-6)^2+(8-6)^2=8 dollars, which is the minimum.

Sample Input 2

3
1 1 3

Sample Output 2

3

Transforming the all into 2s will cost (1-2)^2+(1-2)^2+(3-2)^2=3 dollars. Note that Evi has to pay (1-2)^2 dollar separately for transforming each of the two 1s.

Sample Input 3

3
4 2 5

Sample Output 3

5

Leaving the 4 as it is and transforming the 2 and the 5 into 4s will achieve the total cost of (2-4)^2+(5-4)^2=5 dollars, which is the minimum.

Sample Input 4

4
-100 -100 -100 -100

Sample Output 4

0

Without transforming anything, Evi's objective is already achieved. Thus, the necessary cost is 0.

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

import java.util.*;

public class Main {
	public static void main(String[] args) {
		Scanner sc = new Scanner(System.in);
		int numbers = sc.nextInt();
		int[] values = new int[numbers];
		int sum = 0;
		for (int i = 0; i < numbers; i++) {
			values[i] = sc.nextInt();
			sum += values[i];
		}
		
		// finding the average
		int averageCeil = (int) Math.ceil(sum * 1.0/numbers);
		int averageFloor = (int) Math.floor(sum * 1.0/numbers);
		
		// computing the cost
		int totalcost = 0; 
		for (int i = 0; i < numbers; i++) {
			int x = Math.min(((values[i] - averageCeil) * (values[i] - averageCeil)),(values[i] - averageFloor) * (values[i] - averageFloor));
			totalcost += x;
		}
		System.out.println(totalcost);
	}
}