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<iostream>
#include<vector>
#include<algorithm>
#include<string>
#include<set>
#include<map>
#include<queue>
#define REP(i,a) for (int i = 0;i < (a);++i)
#define FOR(i,a,b) for (int i = (a);i < (b); ++i)
#define FORR(i,a,b) for (int i = (a);i >= (b); --i)
#define ALL(obj) (obj).begin(),(obj).end()
#define SORT(list) sort(ALL((list)));
#define MOD 1000000007
using namespace std;
using ll = long long;
int main(){
    int n,total=0,mean,mean2,mean3;
    cin >> n;
    vector<int>a(n);
    REP(i,n){
        cin >> a[i];
        total += a[i];
    }
    mean = total/n;
    mean2 = mean+1;
    mean3 = mean-1;

    total = 0;
    int answer = 1000000000;
    REP(i,n){
        total+=(a[i]-mean)*(a[i]-mean);
    }
    answer = min(total,answer);
    total = 0;
    REP(i,n){
        total+=(a[i]-mean2)*(a[i]-mean2);
    }
    answer = min(total,answer);
    total = 0;
    REP(i,n){
        total+=(a[i]-mean3)*(a[i]-mean3);
    }
    answer = min(total,answer);
    cout << answer << endl;
    return 0;
}