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 5.4.1) (judge verdict: Wrong Answer)

#include<stdio.h>

int main(){
  //入力を受け取る
  int n,i,j,min=101,max=-101,sum=0,ans=101;
  scanf("%d",&n);
  int a[n];
  for(i=0;i<n;++i){
    scanf("%d",&a[i]);
  }
  //最小値計算,最大値計算
  for(i=0;i<n;++i){
    if(a[i]>max)max=a[i];
    if(a[i]<min)min=a[i];
  }
  //全探索（最小値から最大値までに解は必ずある）
  for(i=min;i<=max;i++){//i=解
    for(j=0;j<n;j++){//解との差の絶対値をそれぞれに対して計算(n回繰り返し)
      sum+=abs(i-a[j])*abs(i-a[j]);
    }
    if(sum<ans){
      ans=sum;
    }
    sum=0;
  }
  printf("%d\n",ans);
  return 0;
}