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>
#include <stdlib.h>
#include <string.h>
#include <math.h>

#define MAX(a,b) (a > b) ? a : b
#define MIN(a,b) (a < b) ? a : b
const int inf = 1000000000; // 10^9

int Amax(int *p,int size){
   int _max = 0;
   for(int i = 0;i < size;i++)
      if(p[i] > _max)_max = p[i];
   return _max;
}
int Amin(int *p,int size){
   int _min = 0;
   for(int i = 0;i < size;i++)
      if(p[i] < _min)_min = p[i];
   return _min;
}
int main(){
   int n;scanf("%d",&n);
   int a[100];
   for(int i = 0;i < n;i++)scanf("%d",&a[i]);
   int ans = 100;
   for(int i = Amin(a,n);i < Amax(a,n);i++){
      int _ans = 0;
      for(int j = 0;j < n;j++){
         _ans += pow(abs(a[j] - i),2);
      }
      //printf("MIN(%d,%d)\n",ans,_ans);
      ans = MIN(ans,_ans); 
      //printf("%d:%d\n",i,ans);
   }
   printf("%d",ans);
}