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 Java (OpenJDK 11.0.6) (judge verdict: Wrong Answer)

import java.util.*;
class Main{
    public static void main(String[]args){
        Scanner scan = new Scanner(System.in);
        int N = scan.nextInt();
        int[] X = new int[N];
        int[] sq = new int[N];

        for(int i=0;i<N;i++){
            X[i] = 0;
            sq[i] = 0;
        }

        for(int i=0;i<N;i++){
            X[i] = scan.nextInt();
            sq[i] = X[i]*X[i];
        }

        double avg = Sum(X, N)/(double)N;
        System.out.println(avg);

        System.out.println((int)Math.ceil(f(X,N)));

        scan.close();
    }

    public static int Sum(int[] x,int N){
        int sum = 0;
        for(int i=0;i<N;i++){
            sum += x[i];
        }

        return sum;
    }

    public static double f(int[]X,int N){
        double val = 0;

        for(int i=0;i<N;i++){
            val += ((double)X[i]-(double)(Sum(X,N))/N)*((double)X[i]-(double)(Sum(X,N))/N);
        }

        return val;
    }

}