raise to the power in binary
2 0 =1-> 1
2 1
=2-> 10
2
2
=4-> 100
we can check the binary forms using bitwise and shift operators
/* number in power of 2 */
#include<stdio.h>
main()
{
int num, b, power, flag;
printf("\n enter number");
scanf("%d", &num);
b=1;
power=0;
flag=0;
while(b<=num)
{
if( (num^b)==0)
{
flag++;
break;
}
power++;
b<<=1;
}
if(flag)
printf("%d is 2 raise to the power %d", num, power);
else
printf("%d is not any number raise to power 2",num);
return 0;
}
suppose i enter 2049 as input the above program takes 11 comparisons to figure out whether its a number is equal to any power of 2 which its not . similarly it take 2 comparisions in case
of 5. so to reduce number of comparisions we can can exploit a property that if a number is equal to any power raise to 2 it will have only one 1 which will be at the beginning of its binary. so here is another code
#include<stdio.h>
main()
{
int num,rem,flag=1, loops=0;
printf("enter a number");
scanf("%d",num);
while(num>1)
{
loops++;
rem=num%2;
num=num/2;
if(rem)
{
flag--;
break;
}
}
if(flag)
printf("yes is power of 2");
else
printf("it is not");
printf("\n loops=%d",loops);
return 0;
}
in above program number of loops also tell the power 2 is raised to get that number if it is.
both the programs do not work if the user enter number <=0
so we can also add a check .. i havent done that though..
