mirror of
https://github.com/TheAlgorithms/Python.git
synced 2024-11-27 23:11:09 +00:00
c6582b35bf
Co-authored-by: pre-commit-ci[bot] <66853113+pre-commit-ci[bot]@users.noreply.github.com> Co-authored-by: Dhruv Manilawala <dhruvmanila@gmail.com> Co-authored-by: Christian Clauss <cclauss@me.com>
78 lines
2.0 KiB
Python
78 lines
2.0 KiB
Python
"""
|
|
Project Euler Problem 64: https://projecteuler.net/problem=64
|
|
|
|
All square roots are periodic when written as continued fractions.
|
|
For example, let us consider sqrt(23).
|
|
It can be seen that the sequence is repeating.
|
|
For conciseness, we use the notation sqrt(23)=[4;(1,3,1,8)],
|
|
to indicate that the block (1,3,1,8) repeats indefinitely.
|
|
Exactly four continued fractions, for N<=13, have an odd period.
|
|
How many continued fractions for N<=10000 have an odd period?
|
|
|
|
References:
|
|
- https://en.wikipedia.org/wiki/Continued_fraction
|
|
"""
|
|
|
|
from math import floor, sqrt
|
|
|
|
|
|
def continuous_fraction_period(n: int) -> int:
|
|
"""
|
|
Returns the continued fraction period of a number n.
|
|
|
|
>>> continuous_fraction_period(2)
|
|
1
|
|
>>> continuous_fraction_period(5)
|
|
1
|
|
>>> continuous_fraction_period(7)
|
|
4
|
|
>>> continuous_fraction_period(11)
|
|
2
|
|
>>> continuous_fraction_period(13)
|
|
5
|
|
"""
|
|
numerator = 0.0
|
|
denominator = 1.0
|
|
root = int(sqrt(n))
|
|
integer_part = root
|
|
period = 0
|
|
while integer_part != 2 * root:
|
|
numerator = denominator * integer_part - numerator
|
|
denominator = (n - numerator**2) / denominator
|
|
integer_part = int((root + numerator) / denominator)
|
|
period += 1
|
|
return period
|
|
|
|
|
|
def solution(n: int = 10000) -> int:
|
|
"""
|
|
Returns the count of numbers <= 10000 with odd periods.
|
|
This function calls continuous_fraction_period for numbers which are
|
|
not perfect squares.
|
|
This is checked in if sr - floor(sr) != 0 statement.
|
|
If an odd period is returned by continuous_fraction_period,
|
|
count_odd_periods is increased by 1.
|
|
|
|
>>> solution(2)
|
|
1
|
|
>>> solution(5)
|
|
2
|
|
>>> solution(7)
|
|
2
|
|
>>> solution(11)
|
|
3
|
|
>>> solution(13)
|
|
4
|
|
"""
|
|
count_odd_periods = 0
|
|
for i in range(2, n + 1):
|
|
sr = sqrt(i)
|
|
if sr - floor(sr) != 0:
|
|
if continuous_fraction_period(i) % 2 == 1:
|
|
count_odd_periods += 1
|
|
return count_odd_periods
|
|
|
|
|
|
if __name__ == "__main__":
|
|
print(f"{solution(int(input().strip()))}")
|