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;

#define vit vector<int>::iterator
#define pb push_back
#define ppb pop_back
#define mp make_pair
#define all(x) (x).begin(), (x).end()
#define forn(i, n) for(int i = 0; i < int(n); i++)
#define fr first
#define sc second
#define skip continue
#define PI 3.14159265

typedef unsigned long long ull;
typedef long long ll;
typedef unsigned int ui;
typedef long double ld;
typedef vector<int> vi;
typedef vector<long long> vll;
typedef vector<bool> vb;

const int INF = 1e9 + 5;
const int MAXN = 3e6 + 123;

void faster(){
    ios_base::sync_with_stdio(false);
    cin.tie(NULL);
    cout.tie(NULL);
}
bool cmp(const string& a, const string& b) {
	return (a + b) < (b + a);
}

vector<int> Reversed(const vector<int>& s){
   vector<int> d = s;
   reverse(all(d));
   return d;
}

void solve(){
    int n;
    cin >> n;
    vi v(n);
    int sum = 0;
    for(int i = 0; i < n; i++){
        cin >> v[i];
        sum += v[i];
    }
    ll cnt = 0, t = sum / n;
    for(int i = 0; i < n; i++){
        cnt += (v[i] - t)*(v[i] - t);
    }
    ll cnt1 = 0;
    t = sum / n + 1;
    for(int i = 0; i < n; i++){
        cnt1 += (v[i] - t)*(v[i] - t);
    }
    cout << min(cnt, cnt1);
}

// ."ww"w"".ww"""...ww"".

int main(){
    //freopen("lepus.in","r",stdin);
    //freopen("lepus.out","w",stdout);
    //faster();
    int n=1;
    //cin>>n;
    while(n--){
    solve();
    //cout << endl;
    }
    //fclose(stdout);
}