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++14 (GCC 5.4.1) (judge verdict: Accepted)

#include<bits/stdc++.h>
using namespace std;

int n,ar[2000];

int f(int i){

              int cost=0;
            for(int j=0;j<=n-1;j++)
                cost+=(int)abs(ar[j]-i)*(int)abs(ar[j]-i);

    return cost;
}

int main(){

    while(cin>>n){

        for(int i=0;i<n;i++)
            scanf("%d",ar+i);

        sort(ar,ar+n);

        int mncost=INT_MAX;
        int low=ar[0],high=ar[n-1];

        if(high-low==0){
            puts("0");
            continue;
        }

        while(high-low>1){
            int mid=(high+low)/2;

            if(f(mid)<f(mid+1))
                high=mid;
            else if(f(mid)>f(mid+1))
                low=mid;

        }

        cout<<f(high)<<endl;

    }




return 0;
}

/*

5
-5 -3 0 2 9
ans=118

5
1 2 3 4 5
ans=10

*/