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 <cstdio>
#include <cstring>
#include <cmath>
#include <algorithm>
#include <sstream>
#include <iostream>
#include <vector>
#include <map>
#include <set>

using namespace std;

typedef unsigned long long ull;
typedef long long ll;
typedef vector<int> vi;
typedef vector<long> vl;

#define POW2(x) ((x) * (x))

int
main()
{
    int N;
    vector<int> a;
    cin >> N;
    for (int i = 0; i < N; ++i) {
        int tmp;
        cin >> tmp;
        a.push_back(tmp);
    }
    const int l = -100, r = 101;
    int min_cost = 200 * 200 * N;
    int min_num = -100;
    for(int i = l; i < r; ++i) {
        int cost = 0;
        for (int j = 0; j < N; ++j) {
            cost += POW2(a[j] - i);
        }
        
        if (cost <= min_cost) {
            min_num = i;
            min_cost = cost;
        }
    }
    cout << min_cost << endl;
    return 0;
}