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>
#include <stdlib.h>
#define INT_MAX 1000000000
//Se você tivesse declarado a biblioteca limits.h não precisaria definir INT_MAX. ela é um macro dessa biblioteca.
int ordena(const void *a, const void *b){
  return (*(int *)a)-(*(int *)b);
}

int main (){    //EXERCICIO DE FORÇA BRUTA.
  int n, a[112];

  scanf("%d", &n);

  for(int i=0; i<n ; i++)
    scanf("%d", &a[i] );

  qsort(a,n,sizeof(int), ordena);

  int min = INT_MAX;
  for(int i=a[0]; i<=a[n-1]; i++){
    int soma=0;
    for(int j=0; j<n; j++){
      soma+= (a[j]-i)*(a[j]-i);
  //    printf("%d ", soma);
    }
    if(min>soma)
      min=soma;
  }
  printf("%d\n", min);

  return 0;
}