After the power of 2 is written in binary form, it is easy to find a feature: there is only one 1 in the binary, and 1 followed by n 0, so the question can be converted to determine whether 1 followed by n 0.
If you subtract this number by 1, you will find that the only 1 will become 0, and the original n 0 will be 1, so that the original number and the number after subtracting 1 will be zero.
The quickest way:
(number & number-1) = = 0
Reason:Because 2 of the n-th conversion is binary to 10 ... 0 such a form (excluding 0). With the digits on their own-1, this gets the result of 0. For example. The binary of 8 is the binary of 1000;8-1=7,7 is 111. The result of the two phase is 0. The calculations are as follows:
1000
& 0111
-------
0000
the code that uses the recursive implementation is as follows:
Copy Code code as follows:
#include "stdio.h"
#include "Stdlib.h"
int log2 (int value)//recursive judge how many times a number is 2
{
if (value = = 1)
return 0;
Else
Return 1+log2 (value>>1);
}
int main (void)
{
int num;
printf ("Please enter an integer:");
scanf ("%d", &num);
if (num& (num-1))//use and operation to determine whether a number is a power exponent of 2
printf ("%d is not a power party of 2!") \ n ", num);
Else
printf ("%d is 2%d Times Square!") \ n ", num,log2 (num));
System ("pause");
return 0;
}
The code that uses the non-recursive implementation is as follows:
Copy Code code as follows:
#include "stdio.h"
#include "Stdlib.h"
int log2 (int value)//non-recursive judge how many times a number is 2
{
int x=0;
while (value>1)
{
value>>=1;
x + +;
}
return x;
}
int main (void)
{
int num;
printf ("Please enter an integer:");
scanf ("%d", &num);
if (num& (num-1))//use and operation to determine whether a number is a power exponent of 2
printf ("%d is not a power party of 2!") \ n ", num);
Else
printf ("%d is 2%d Times Square!") \ n ", num,log2 (num));
System ("pause");
return 0;
}
Extension: Find the number of 1 in binary of N.
A very ingenious use of a property,n=n& (n-1) can remove the property of the rightmost 1 in the binary of N, and the loop is removed until all 1 is removed, which reduces the complexity of the problem to only the number of 1. The code is as follows:
Copy Code code as follows:
int Func3 (int data)
{///using data& (data-1) can remove the rightmost 1 each time, remove how many 1, that is, contains a few 1
int count = 0;
while (data)
{
data = data & (DATA-1);
count++;
}
return count;
}
Extended Question two:
How many bits are different in the binary of A and B. This problem can be divided into two steps, (1) different from A and B or get C, i.e. C=a^b, (2) to calculate how many 1 in the C binary.
