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


int sift(int l, int r);

int main(){

	int *a, *b;
	char buf[256];
	int i, j, k, n, sumb, ret, ch;
	char* s;
	int ball=0;
	int binb=1;
	
	if(fgets(buf, 256, stdin) == NULL){
		
		return 1;
		
	}
	
	n=atoi(buf);
	
	a=(int *)malloc(sizeof(int)*n);
	b=(int *)malloc(sizeof(int)*n);
	
	i=0;
	
	while((ch = getchar()) != EOF){
		
		if(ch != 32 && ch != 10){a[i++]=(ch-48);}
				
	}
	
	for(i=0; i < n; i++){
		
		b[i]=0;
		
	}
	
	for(j=n-1; j >= 0; j--){
		
		if(j > (n/2 - 1)){b[j]=a[j];}
		else{
			
			sumb=0;
			
			for(k = ((n/(j+1)-1)); k > 1; k--){
				
				sumb=sumb+b[k*(j+1)-1];
				
			}
			
			if(sumb % 2 == a[j]){b[j]=0;}
			else{b[j]=1;}
			
		}
		
	}
	
	for(i=0; i < n; i++){
		
		ball=ball+b[i];
		
	}
	
	printf("%d\n", ball);
	
	for(i=0; i < n; i++){
		
		if(b[i] == 1){
		
		if(binb == ball){printf("%d\n", i+1);}
		else{
			printf("%d ", i+1);
			binb++;
		}
			
		}
		
	}
	
	free(a);
	free(b);
	
	
}