sol1.py

"""
Project Euler Problem 57: https://projecteuler.net/problem=57
It is possible to show that the square root of two can be expressed as an infinite
continued fraction.

sqrt(2) = 1 + 1 / (2 + 1 / (2 + 1 / (2 + ...)))

By expanding this for the first four iterations, we get:
1 + 1 / 2 = 3 / 2 = 1.5
1 + 1 / (2 + 1 / 2} = 7 / 5 = 1.4
1 + 1 / (2 + 1 / (2 + 1 / 2)) = 17 / 12 = 1.41666...
1 + 1 / (2 + 1 / (2 + 1 / (2 + 1 / 2))) = 41/ 29 = 1.41379...

The next three expansions are 99/70, 239/169, and 577/408, but the eighth expansion,
1393/985, is the first example where the number of digits in the numerator exceeds
the number of digits in the denominator.

In the first one-thousand expansions, how many fractions contain a numerator with
more digits than the denominator?
"""


defsolution(n: int=1000) ->int:
"""
 returns number of fractions containing a numerator with more digits than
 the denominator in the first n expansions.
 >>> solution(14)
 2
 >>> solution(100)
 15
 >>> solution(10000)
 1508
 """
prev_numerator, prev_denominator=1, 1
result= []
foriinrange(1, n+1):
numerator=prev_numerator+2*prev_denominator
denominator=prev_numerator+prev_denominator
iflen(str(numerator)) >len(str(denominator)):
result.append(i)
prev_numerator=numerator
prev_denominator=denominator

returnlen(result)


if__name__=="__main__":
print(f"{solution() =}")