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 C (GCC 5.4.1) (judge verdict: Accepted)

#include <stdio.h>

int main(void) {
	int n, sum = 0, ave, max, min;
	long long total = 0;
	int a[100];
	int i, j;
	int mode, cnt, temp_cnt;
	
	scanf("%d", &n);
	
	for (i = 0; i < n; i++){
		scanf("%d", &a[i]);
		if (i == 0){
			min = a[i];
			max = a[i];
		}
		if (a[i] < min) min = a[i];
		if (a[i] > max) max = a[i];
		sum += a[i];
	}
	
	mode = 0;
	cnt = 0;
	
	for (i = 0; i < n; i++){
		temp_cnt = 1;
		for (j = i + 1; j < n; j++){
			if (a[i] == a[j]) temp_cnt++;
		}
		if (temp_cnt > cnt){
			cnt = temp_cnt;
			mode = a[i];
		}
	}
	
	if (sum % n == 0) ave = sum/n;
	else{
		if ((mode < ave) && (cnt > max)) ave = sum/n;
		else ave = (sum+(n/2))/n;
	}
	
	for (i = 0; i < n; i++){
		total += (a[i] - ave)*(a[i] - ave);
	}
	
	printf("%ld\n", total);
	
	return 0;
}