Problem:
Score : 300 points

Problem Statement

There are N people living on a number line.

The i-th person lives at coordinate X_i.

You are going to hold a meeting that all N people have to attend.

The meeting can be held at any integer coordinate. If you choose to hold the meeting at coordinate P, the i-th person will spend (X_i - P)^2 points of stamina to attend the meeting.

Find the minimum total points of stamina the N people have to spend.

Constraints

All values in input are integers.

1 \leq N \leq 100

1 \leq X_i \leq 100

Input

Input is given from Standard Input in the following format:

N
X_1 X_2 ... X_N

Output

Print the minimum total stamina the N people have to spend.

Sample Input 1

2
1 4

Sample Output 1

5

Assume the meeting is held at coordinate 2. In this case, the first person will spend (1 - 2)^2 points of stamina, and the second person will spend (4 - 2)^2 = 4 points of stamina, for a total of 5 points of stamina. This is the minimum total stamina that the 2 people have to spend.

Note that you can hold the meeting only at an integer coordinate.

Sample Input 2

7
14 14 2 13 56 2 37

Sample Output 2

2354

Solution in C++14 (GCC 5.4.1) (judge verdict: Accepted)

#include <bits/stdc++.h>
typedef long long ll;
typedef long double ld;

using namespace std;
const ll LINF = LLONG_MAX;
const int INF = INT_MAX;
const int MOD = 998244353;

template <typename T>
void read(vector<T>& a) {
    for (auto &v : a)
        cin >> v;
}

int a[100];
int n;

ll test(int p){
    ll ans = 0;
    for (int i= 0; i< n; i++)
        ans += 1ll * (a[i] - p) * (a[i] - p);
    return ans;
}

int main() {
    ios::sync_with_stdio(false);
    cin.tie();
    cin >> n;
    ll s = 0;
    for (int i = 0; i < n; i++){
        cin >> a[i];
        s += a[i];
    }
    int m = ld(s)/n;
    ll ans = INF;
    for (int i = m-1; i <= m + 1; i++)
        ans = min(ans, test(i));
    cout << ans;
}