Problem:
Score : 100 points

Problem Statement

It is known that the area of a regular dodecagon inscribed in a circle of radius a is 3a^2.

Given an integer r, find the area of a regular dodecagon inscribed in a circle of radius r.

Constraints

1 \leq r \leq 100

r is an integer.

Input

Input is given from Standard Input in the following format:

r

Output

Print an integer representing the area of the regular dodecagon.

Sample Input 1

4

Sample Output 1

48

The area of the regular dodecagon is 3 \times 4^2 = 48.

Sample Input 2

15

Sample Output 2

675

Sample Input 3

80

Sample Output 3

19200

Solution in C (GCC 5.4.1) (judge verdict: Wrong Answer)

#include <stdio.h>
int main(void){
  int n;
  scanf("%d", &n);
  int a[n];
  int b[n];
  int n_right = -1;

  for(int i=0;i<n;i++) {
  	scanf("%d", &a[i]);
  }

  int s;
  for(int i=n-1;i>=0;i--){
    b[i] = 0;
    s = 0;
    //s: kがiの倍数のb[k]の和
    //a_2 b_2,4,6,8
    //i:1 k:1,3,5,7 k+=(i+1)
    //a_1 b_1,2,3,4
    //i:0 k:0,1,2,3 k+=(i+1)

    for(int k=i; k<n; k+=i+1){
      s += b[k];
    }
    unsigned int tmp_s = s&1;
    unsigned int tmp_a = a[i]&1;
    if(tmp_s != tmp_a){
    	b[i] = 1;
    	if(n_right<0) n_right = i;
    } 
  }
  
  if(n_right<0){
    printf("0");  	
  }else if(n_right==0){
  	printf("%d", b[0]);
  }else{
  	for(int i=0; i<=n_right; i++){
  		printf("%d ", b[i]);
    }
  }
  


    
  return 0;
}