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<iostream>
#include<algorithm>
#include<vector>
#define lamba ios::sync_with_stdio(0), cin.tie(0), cout.tie(0)
#define all(x) x.begin(), x.end()
using namespace std;

const int zyan = 1e2 + 1;

int n, a[105], mn = zyan, mx = -zyan;

int fast() {
cin >> n;

for(int i=1;i<=n;i++){
cin >> a[i];
mn = min(mn, a[i]);
mx = max(mx, a[i]);
}
/*
int checker = 0;

for(int i=2;i<=n;i++){
	if(a[i] == a[1])checker++;
}


if(checker == n-1){
cout << 0;
return 0;
}*/

int mid = (mn + mx) % 2 + (mn + mx) >> 1;

int res = 0;

for(int i=1;i<=n;i++){
	if(mid == a[i])continue;
	else res += (a[i] * a[i]) - (a[i] * mid) - (mid * a[i]) + (mid * mid);
}

cout << res;

}

int main(){
lamba;

	cin >> n;
	
	for (int i = 1; i <= n; ++i) {
		cin >> a[i];
	}

	int ans = zyan;

	for (int j = -100; j <= 100; ++j) {
		int cur = 0;
		for (int i = 1; i <= n; ++i) {
			cur += (a[i] - j) * (a[i] - j);
		}
		ans = min(ans, cur);
	}
	
	cout << ans;
return 0;
}