Problem:
Score : 200 points

Problem Statement

Evi has N integers a_1,a_2,..,a_N. His objective is to have N equal integers by transforming some of them.

He may transform each integer at most once. Transforming an integer x into another integer y costs him (x-y)^2 dollars. Even if a_i=a_j (i≠j), he has to pay the cost separately for transforming each of them (See Sample 2).

Find the minimum total cost to achieve his objective.

Constraints

1≦N≦100

-100≦a_i≦100

Input

The input is given from Standard Input in the following format:

N
a_1 a_2 ... a_N

Output

Print the minimum total cost to achieve Evi's objective.

Sample Input 1

2
4 8

Sample Output 1

8

Transforming the both into 6s will cost (4-6)^2+(8-6)^2=8 dollars, which is the minimum.

Sample Input 2

3
1 1 3

Sample Output 2

3

Transforming the all into 2s will cost (1-2)^2+(1-2)^2+(3-2)^2=3 dollars. Note that Evi has to pay (1-2)^2 dollar separately for transforming each of the two 1s.

Sample Input 3

3
4 2 5

Sample Output 3

5

Leaving the 4 as it is and transforming the 2 and the 5 into 4s will achieve the total cost of (2-4)^2+(5-4)^2=5 dollars, which is the minimum.

Sample Input 4

4
-100 -100 -100 -100

Sample Output 4

0

Without transforming anything, Evi's objective is already achieved. Thus, the necessary cost is 0.

Solution in C (GCC 9.2.1) (judge verdict: Accepted)

#include<stdio.h>
#include<stdlib.h>
#define kai(a) (a)*(a)
int min(int a[],int n){
    int ret;
    ret=a[0];
    for(int i=1;i<n;i++){
        if(a[i]<=ret) ret=a[i];
    }
    return ret;
}
int max(int a[], int n){
    int ret=a[0];
    for(int i=1;i<n;i++){
        if(a[i]>=ret) ret=a[i];
    }
    return ret;
}

int main(){
    int n;
    int *a;
    int *sum;
    scanf("%d",&n);
    a=calloc(n,sizeof(int));
    sum=calloc(max(a,n)-min(a,n)+1,sizeof(int));
    for(int i=0;i<n;i++){
        scanf("%d",&a[i]);
    }
    int j=0;
    for(int i=min(a,n);i<=max(a,n);i++){
        int ssum=0;
        for(int k=0;k<n;k++){
            ssum+=kai(a[k]-i);
        }
        sum[j++]=ssum;
    }


    printf("%d",min(sum,max(a,n)-min(a,n)+1));

    printf("\n");
    return 0;
}