human
Can you provide a code to count numbers whose sum with x is equal to XOR with x and the other steps required given the following:
Count numbers whose sum with x is equal to XOR with x | Function to find total 0 bit in a number.
Function to find Count of non - negative numbers less than or equal to x , whose bitwise XOR and SUM with x are equal ..
count number of zero bit in x.
power of 2 to count.
Driver code.
Function call"
How can the above be solved in Python?
assistant
#Here is code to count numbers whose sum with x is equal to XOR with x and the other steps required.
 def CountZeroBit ( x ) :
	count = 0
	while ( x ) :
		if ( ( x & 1 ) == 0 ) :
			count += 1
		x >>= 1
	return count
def CountXORandSumEqual ( x ) :
	count = CountZeroBit ( x )
	return ( 1 << count )
if __name__ == ' _ _ main _ _ ' :
	x = 10
	print ( CountXORandSumEqual ( x ) )