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 (Clang 3.8.0) (judge verdict: Wrong Answer)

#include <algorithm>
#include <bitset>
#include <cstdio>
#include <cassert>
#include <cmath>
#include <functional>
#include <iostream>
#include <map>
#include <stack>
#include <queue>
#include <set>
#include <tuple>
#include <utility>
#include <vector>
using namespace std;

typedef long long LL;
typedef pair<int, int> P;
typedef pair<LL, LL> PLL;

int main() {
    int N;
    int a[105];

    cin >> N;
    for (int j = 0; j < N; ++j) {
        cin >> a[j];
    }

    int ret[200];

    fill(ret, ret + 200, 0);
    for (int j = 0; j <= 200; ++j) {
      for (int k = 0; k < N; ++k) {
        int offset = a[k] - (j - 100);
        ret[j] += offset * offset;
      }
    }

    int ans = 100000000;
    for (int j = 0; j < 200; ++j) {
        ans = min(ans, ret[j]);
    }

    cout << ans << endl;
    return 0;
}