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>

int main(void) {
    int N;
    int a[100];
    int i,cnt;
    char str[512];
    char *p;

    int rd=0,ru=0,minus=0 ;
    int rdchk=0,ruchk=0,minux=0;
    
    fgets(str, 512, stdin);
    N = atoi(str);
    fgets(str, 512, stdin);
    p = str;
    for(i=0,cnt=0;i<512;i++){
        if(str[i] == ' '){
            str[i] = '\0';
            a[cnt++] = atoi(p);
            p = &str[i+1];
        }
    }
    a[cnt] = atoi(p);

    for(cnt=0;cnt<N;cnt++){
        rd +=a[cnt];
    }

    rd = rd / N;
    ru = rd + 1;
    minus = rd - 1;

    for(cnt=0;cnt<N;cnt++){
        rdchk += (a[cnt] - rd) * (a[cnt] - rd);
        ruchk += (a[cnt] - ru) * (a[cnt] - ru);
        minux += (a[cnt] - minus) * (a[cnt]-minus);
    }
    
    if(rdchk > ruchk) {
        printf("%d\n", ruchk);
    } else {
        if(ruchk > minux) {
        printf("%d\n", minux);
        } else {
        printf("%d\n", rdchk);
        }
    }

    return 0;
    
}