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>

#define ll long long
#define rep(i, n) for(int i = 0; i < n; i++)
#define rrep(i, n) for(int i = 1; i <= n; i++)
#define drep(i, n) for(int i = n-1; i >= 0; i--)
#define srep(i, s, t) for (int i = s; i < t; i++)
#define vi vector<int>
#define P pair<ll, ll>
using namespace std;
int mod = 1'000'000'000 + 7;
struct edge {
    int to;
    ll cost;
};

void COME_LET_THE_GAMES_BEGIN() {
    cin.tie(nullptr);
    ios::sync_with_stdio(false);
    cout << setprecision(20);
}

int main() {
    COME_LET_THE_GAMES_BEGIN();
    ll N, a[100];
    cin >> N;
    rep(i, N)cin >> a[i];
    ll ans = LLONG_MAX;
    for (int i = -100; i <= 100; i++) {
        ll r = 0;
        rep(j, N) {
            r += pow(a[j] - i, 2);
        }
        ans = min(ans, r);
    }
    cout << ans << endl;
}