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: Wrong Answer)

#include<stdio.h>
#include<math.h>

int main()
{
    long long int n,i,x=0,ave,f=1,y;
    scanf("%lld",&n);
    long long int a[n],sum=0;
    scanf("%lld",&a[1]);
    y=a[1];
    sum=a[1];

    for(i=2; i<=n; i++)
    {
        scanf("%lld",&a[i]);

        sum=sum+a[i];
    }
    for(i=1; i<=n; i++)
    {
        if(a[i]==a[i+1])
            f=0;
        else
            f=1;
    }

    ave=(sum/n);

    if(sum%n!=0 && sum>0)
        ave=ave+1;
    if(sum%n!=0 && sum<0)
            ave=ave-1;


    for(i=1; i<=n; i++)
    {
        x=x+(a[i]-ave)*(a[i]-ave);
    }
    if(f==0)
        printf("0\n");
    else
        printf("%lld\n",x);

    return 0;
}