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: Wrong Answer)

#include <stdio.h>
#include <stdlib.h>
#include <string.h>
#include <math.h>

void swap (int *x, int *y) {
  int temp;

  temp = *x;
  *x = *y;
  *y = temp;
}

void bubble_sort (int array[], int array_size) {
  int i, j;

  for (i = 0; i < array_size - 1; i++){
    for (j = array_size - 1; j >= i + 1; j--){
      if (array[j] < array[j-1]) { swap(&array[j], &array[j-1]); }
    }
  }
}

int main(){

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

    bubble_sort(a,n);

    int tmp, j;
    long int answer = INFINITY;
    for(i = a[0]; i <= a[n-1] ; i++){
        tmp = 0;
        for(j = 0; j < n; j++){
            tmp += pow(a[j]-i,2);
        }
        printf("i = %d, tmp = %d\n",i,tmp);
        if(tmp <= answer){
            answer = tmp;      
        }
        printf("i = %d, answer = %ld\n",i,answer);
    }
    printf("%ld",answer);
   return 0;
}