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: Wrong Answer)

#include <bits/stdc++.h>
         
typedef long long ll;
using namespace std;
        
#define F first
#define S second
#define sqr(x) x*x
#define pb push_back
#define mp make_pair
#define sz(x) ((int) (x).size())
#define all(x) (x).begin(), (x).end()
#define allr(x) (x).rbegin(), (x).rend()
#define repit(i, n) for(auto it = n.begin(); it != n.end(); ++it)
       
const ll mod=1000000007;
ll powmod(ll a,ll b) {ll res=1;a%=mod; assert(b>=0);for(;b;b>>=1){if(b&1)res=res*a%mod;a=a*a%mod;}return res;}
ll gcd(ll a,ll b){if(b==0)return a;else return gcd(b,a%b);}
int lcm(int a,int b){return abs(a*b)/gcd(a,b);}
     
int main(){ios_base::sync_with_stdio(0);cin.tie(0);
    int n;
    cin >> n; 
    int arr[n];
    double s = 0;
    int s1 = 0;
    for(int i = 0; i < n; ++i){
        cin >> arr[i];
        s += arr[i];
        s1 += arr[i];
    }
    s = ceil(s / n);
    int s2 = int(s);
    s1 = s1 / n;
    int ans = 0;
    int ans1 = 0;
    for(int i = 0; i < n; ++i){
        ans += pow((arr[i] - s1),2);
        ans1 += pow((arr[i] - s2),2);
    }
    cout << min(s2, s1);
    
}