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 tmp;
  int N;
  int p;
  int i,k = 0;
  scanf("%d/n",&N);
  int K = N;
  int X[N];
  int Y[101];
  while (i < N)
  {
    scanf("%d ",&X[i]);
    i++;
  }
  i = 0;
  while (i < N)
  {
    k = i + 1;
    while (k < N)
    {
      if (X[i] > X[k])
      {
        tmp = X[k];
        X[k] = X[i];
        X[i] = tmp;
      }
      k++;
    }
    i++;
  }
  i = 0;
  k = 0;
  while (i < N)
  {
    //printf("%d\n",X[i]);
    i++;
  }
  K = X[0];
  p = 0;
  while (K <= X[N - 1])
  {
    i = 0;
    Y[p] = 0;
  	while (i < N)
  	{
    	Y[p] += (X[i] - K) * (X[i] - K);
      	//printf("%d\n",Y[K]);
    	i++;
  	}
    //printf("%d\n",Y[p]);
    p++;
    K++;
  }
  i = 0;
  while (i < X[N-1] - X[0])
  {
    k = i + 1;
    while (k < X[N-1] - X[0])
    {
      if (Y[i] > Y[k])
      {
        tmp = Y[k];
        Y[k] = Y[i];
        Y[i] = tmp;
      }
      k++;
    }
    i++;
  }
  printf("%d",Y[0]);
}