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: Accepted)


import java.io.BufferedReader;
import java.io.InputStreamReader;
import java.util.Arrays;

class Main {
	public static void main(String[] args) throws java.lang.Exception {
		BufferedReader br = new BufferedReader(new InputStreamReader(System.in));
		int n = Integer.parseInt(br.readLine());
		String[] arg = br.readLine().split(" ");
		int[] inp = new int[arg.length];
		for (int i = 0; i < arg.length; i++) {
			inp[i] = Integer.parseInt(arg[i]);
		}
		Solver s = new Solver();

		System.out.println(s.solve(inp));
	}

}

class Solver {

	int solve(int[] inp) {

		int max = Arrays.stream(inp).max().getAsInt();
		int min = Arrays.stream(inp).min().getAsInt();
		int sum = 0;
		int temp = 0;
		for (int i = min; i <= max; i++) {
			temp = 0;
			for (int j = 0; j < inp.length; j++) {
				temp += (i - inp[j]) * (i - inp[j]);
			}
			if (sum > temp||sum==0)
				sum = temp;
		}
		return sum;

	}

}