human
Count of numbers whose sum of increasing powers of digits is equal to the number itself | Function to return the count of digits of n.
Function to return the sum of increasing powers of N.
To store the required answer.
Count of digits in n which will be the power of the last digit.
While there are digits left.
Get the last digit.
Add the last digit after raising it to the required power.
Decrement the power for the previous digit.
Remove the last digit.
Function to return the count of integers which satisfy the given conditions.
If current element satisfies the given condition.
Driver code"
How can the above be solved in Python?

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Can you provide a code to count of numbers whose sum of increasing powers of digits is equal to the number itself and the other steps required?
assistant
#Here is code to count of numbers whose sum of increasing powers of digits is equal to the number itself and the other steps required.
 def countDigits ( n ) :
	cnt = 0
	while ( n > 0 ) :
		cnt += 1
		n //= 10
	return cnt
def digitPowSum ( n ) :
	sum = 0
	pw = countDigits ( n )
	while ( n > 0 ) :
		d = n % 10
		sum += pow ( d , pw )
		pw -= 1
		n //= 10
	return sum
def countNum ( n ) :
	count = 0
	for i in range ( n + 1 ) :
		if ( i == digitPowSum ( i ) ) :
			count += 1
	return count
n = 200
print ( countNum ( n ) )