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 (GCC 5.4.1) (judge verdict: Accepted)

#include <stdio.h>

int main() {
    int n;
    scanf("%d", &n);

    int x[n];

    for(int i = 0; i < n; ++i) // leitura do vetor
      scanf("%d", &x[i]);

    int min = x[0], max = x[0];  /*int min = 101, max = 0;
    declarar os vetores minimos e maximos padrões é um bom exercicio
    mental. */

    for(int i = 1; i < n; ++i) { // encontrando os valores minimos e maximos
      if (x[i] > max)
        max = x[i];
      if (x[i] < min)
        min = x[i];
    }

    int melhortotal;

    for(int p = min; p <= max; ++p) { //calculando o melhor gasto de estamina
      int total = 0;
      for(int i = 0; i < n; ++i) // calculando gasto de estamina
        total += (x[i] - p)*(x[i] - p);
      if (p == min || total < melhortotal) // p == min, garante que o primeiro valor será um que se encontra na lista e não um valor aleatorio.
        melhortotal = total;
    }
    printf("%d\n", melhortotal );

    return 0;
  }
/*   5 / 2 = 2
     5.0 / 2 = 2.5

     int a;

     a = 2.5 -> a = 2   */