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 (Clang 3.8.0) (judge verdict: Wrong Answer)

#include<stdio.h>

int check(int str[],int a){
  for(int i=0;i<a-1;++i){
    if (str[i]!=str[i+1]) {
      return 1;
    }
  }
  return 0;
}

int main(){
  int N=0;
  int sum=0;
  int ave=0;
  int ch=0;
  int min=10000;
  int cost=0;
  int str[100]={};
  scanf("%d",&N);
  for(int i=0;i<N;++i){
    scanf("%d",&str[i]);
  }
  ch=check(str,N);
  if(ch==0){
    printf("0");
  }
  else{
    for(int i=-100;i<=100;++i){
     for(int j=0;j<N;++j){
       cost+=(str[j]-i)*(str[j]-i);
     }
     if (cost<min) {
       min=cost;
     }
     cost=0;
    }
    printf("%d",min);
}
  return 0;
}