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++14 (GCC 5.4.1) (judge verdict: Accepted)

#include<bits/stdc++.h>
using namespace std;
#define ll long long
#define ub upper_bound
//#define int long long
#define vbe(sdk) sdk.begin(),sdk.end()
#define pb push_back
#define loop(i,a,b) for(int i=a;i<b;++i)
#define rloop(i,a,b) for(int i=a;i>=b;--i)
#define mp make_pair
#define mod 1000000007
#define endl '\n'
#define vvll vector<vector<int> >
#define vll vector<int>
#define mint map<int,int> 
#define sz(x) (int)x.size()
#define sll set<int>
#define pll pair<int,int>
#define F first
#define S second
#define pi 3.14159265358979323846
using namespace std;
int arr1[26],arr2[26][26];
void solve()
{
   int n;
   cin>>n;
   double a[n];
   double s=0;
   loop(i,0,n)
   {
       cin>>a[i];
       s+=a[i];
   }
   s=s/n;
   int x=ceil(s);
   int y=floor(s);
  // cout<<x<<y<<"\n";
  int s1=0,s2=0;
  loop(i,0,n)
  {
      s1+=((x-a[i])*(x-a[i]));
  }
  loop(i,0,n)
  {
      s2+=((y-a[i])*(y-a[i]));
  }
  if(s1<s2)
  {
      cout<<s1<<"\n";
  }
  else
  {
      cout<<s2<<"\n";
  }
  
}

signed main(){
  ios_base::sync_with_stdio(false);cin.tie(NULL);cout.tie(NULL);
 
        //  int t;
        // cin>>t;
        // while(t--){
            solve();
    //    }
    return 0;
}