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 "time.h"
#include "unistd.h"
#include "string.h"
#include <math.h>
int ans=0;
int main(int argc, char* argv[]){
  long a=0,b=0,c=100000;
  scanf("%ld",&a);
  long d[a];
  for (int i = 0; i < a; i++) {
    scanf("%ld",&d[i]);
  }
  long min=10000,max=-1;
  for (int i = 0; i < a; i++) {
    if (d[i]<min) {
      min=d[i];
    }
    if (d[i]>max) {
      max=d[i];
    }
  }
  b=max+min;
  b=b/2;
  for (int j = min; j < max; j++) {
    for (int i = 0; i < a; i++) {
      ans += pow(d[i]-j,2);
    }
    if (ans<c) {
      c=ans;
    }
    ans=0;
  }
  printf("%ld",c);
}