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Solution for Euler Problem 26 (#1939)

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* Solution for Euler Problem 26

* Update project_euler/problem_26/sol1.py

typo error fix.

Co-authored-by: Christian Clauss <cclauss@me.com>

* Update project_euler/problem_26/sol1.py

typo error fix

Co-authored-by: Christian Clauss <cclauss@me.com>

* Update project_euler/problem_26/sol1.py

ok to remove, this comes from Pycharm automatically when docstring is added.

Co-authored-by: Christian Clauss <cclauss@me.com>

* Update project_euler/problem_26/sol1.py

ok to remove.

Co-authored-by: Christian Clauss <cclauss@me.com>

* Update project_euler/problem_26/sol1.py

Co-authored-by: Christian Clauss <cclauss@me.com>

* Update project_euler/problem_26/sol1.py

Co-authored-by: Christian Clauss <cclauss@me.com>

* Update project_euler/problem_26/sol1.py

Co-authored-by: Christian Clauss <cclauss@me.com>

* now_divide = now_divide * 10 % divide_by_number

Co-authored-by: Christian Clauss <cclauss@me.com>
This commit is contained in:
Alok Shukla 2020-05-04 02:18:16 +05:30 committed by GitHub
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commit 9b2d65bac1
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project_euler/problem_26/__init__.py Normal file
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"""
Euler Problem 26
https://projecteuler.net/problem=26
Find the value of d < 1000 for which 1/d contains the longest recurring cycle
in its decimal fraction part.
"""
def find_digit(numerator: int, digit: int) -> int:
"""
Considering any range can be provided,
because as per the problem, the digit d < 1000
>>> find_digit(1, 10)
7
>>> find_digit(10, 100)
97
>>> find_digit(10, 1000)
983
"""
the_digit = 1
longest_list_length = 0
for divide_by_number in range(numerator, digit + 1):
has_been_divided = []
now_divide = numerator
for division_cycle in range(1, digit + 1):
if now_divide in has_been_divided:
if longest_list_length < len(has_been_divided):
longest_list_length = len(has_been_divided)
the_digit = divide_by_number
else:
has_been_divided.append(now_divide)
now_divide = now_divide * 10 % divide_by_number
return the_digit
# Tests
if __name__ == "__main__":
import doctest
doctest.testmod()