mirror of
https://github.com/TheAlgorithms/Python.git
synced 2024-12-25 04:30:15 +00:00
49 lines
1.5 KiB
Python
49 lines
1.5 KiB
Python
|
"""
|
||
|
|
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?
|
||
|
|
"""
|
||
|
|
|
||
|
|
|
||
|
|
def solution(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 = []
|
||
|
|
for i in range(1, n + 1):
|
||
|
|
numerator = prev_numerator + 2 * prev_denominator
|
||
|
|
denominator = prev_numerator + prev_denominator
|
||
|
|
if len(str(numerator)) > len(str(denominator)):
|
||
|
|
result.append(i)
|
||
|
|
prev_numerator = numerator
|
||
|
|
prev_denominator = denominator
|
||
|
|
|
||
|
|
return len(result)
|
||
|
|
|
||
|
|
|
||
|
|
if __name__ == "__main__":
|
||
|
|
print(f"{solution() = }")
|