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 9.2.1) (judge verdict: Accepted)

#include <stdio.h>
const int inf = 1000000000; // 10^9
int main(void){
    int n;
    int x[100+5];
    scanf("%d", &n);
    for(int i=0; i<n; i++)scanf("%d", &x[i]);
    
    // 最小値から最大値の範囲の総当り
    int min=x[0], max=x[0];
    for(int i=1; i<n; i++){
        if(min>x[i])min=x[i];
        if(max<x[i])max=x[i];
    }
    
    int ans=inf;
    for(int i=min; i<=max; i++){
        int cost=0;
        for(int j=0; j<n; j++){
            cost+=(x[j]-i)*(x[j]-i);
        }
        if(ans>cost)ans=cost;
    }
    printf("%d\n", ans);
    return 0;
}