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>
#include <math.h>
#define _GLIBCXX_DEBUG
#define _LIBCPP_DEBUG 0

using namespace std;
#define ll long long
#define rep(i,n) for (int i = 0; i < n; i++)
#define MOD (1000000007)
#define vi vector<int>
#define vvi vector<vi>
#define pii pair<int, int>
#define pb push_back
#define mp make_pair
#define all(a) (a).begin(),(a).end()

template<class T> inline bool chmax(T& a, T b) { if (a < b) { a = b; return true; } return false; }
template<class T> inline bool chmin(T& a, T b) { if (a > b) { a = b; return true; } return false; }

int gcd(int a, int b) {
    if (b == 0) return a;
    else return gcd(b, a % b);
}

const ll INF = 1LL << 60;

int main(){
  int n; cin >> n;
  vi a(n); rep(i, n) cin >> a[i];
  int sum = 0; rep(i, n) sum += a[i];

  int cost1 = 0, cost2 = 0;
  // cost を最小にするようなxは実数なら微分法より x = sum÷n(実数) と分かる
  // しかし許されるのは整数だからminを与えるのは sum/n(切り捨てに注意) か sum/n + 1 のいずれか
  //両方試せばよい
  rep(i, n){
    cost1 += (a[i] - sum/n)*(a[i] - sum/n);
    cost2 += (a[i] - sum/n - 1)*(a[i] - sum/n - 1);
  }
  int ans = min(cost1,cost2);

  std::cout << ans << '\n';

}