#include <stdio.h>

/* T(n)=T(n-1)+c => O(n)*/
float stepen1(float x, int a){
	if(a==0)
		return 1;
	return x*stepen1(x,(a-1));
}

float stepen2(float x, int a){
	float t;
	if(a==0)
		return 1;
	if(a%2==0){
		t=stepen2(x,a/2);
		return t*t;	
	}else{
		return x*stepen2(x,a-1);
	}
}

/* 	T(n) = 2*T(n/2)+c 
   	prema master teoremi a=2, b=2, k=0 u formi T(n)=a*T(n/b)+c*n^k
	pa je slozenost O(n^(log_b(a)))=O(n^1)=O(n), znaci nema poboljsanja
*/
float stepen3(float x, int a){
	if(a==0)
		return 1;
	if(a%2)
		return stepen3(x,a/2)*stepen3(x,a/2)*x;
	else
		return stepen3(x,a/2)*stepen3(x,a/2);
}

/* T(n)=T(n/2)+c
   T(n/2)=T(n/4)+c
  ..
   T(1)=T(0)+c
   T(0)=c
----------
+
 T(n)=log_2(n)*c=O(log(n))
*/

float stepen4(float x, int a){
	float t;
        if(a==0)
                return 1;     
        t=stepen4(x,a/2); /* celobrojno deljenje a/2 */
	if(a%2) 
                return t*t*x;
        else
                return t*t;
}


int main(){
	printf("%f %f %f %f\n",stepen1(2,11),stepen2(2,11),stepen3(2,11),stepen4(2,11));
	return 0;
}