mirror of
https://github.com/TheAlgorithms/Python.git
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ef53bbdf5a
* add typehints and docstrings * add typehint and default value * add typehint and default value. Removed unused variable. * do not modifiy the given solution * add doctests * update sol1 after running black * add typehint, docstring, and doctest * update sol2 after running black * add full problem statement and solution function with typehint and doctest * renamed original function instead of adding new one * don't alter original solution
61 lines
1.5 KiB
Python
61 lines
1.5 KiB
Python
"""
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Euler Problem 26
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https://projecteuler.net/problem=26
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Problem Statement:
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A unit fraction contains 1 in the numerator. The decimal representation of the
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unit fractions with denominators 2 to 10 are given:
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1/2 = 0.5
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1/3 = 0.(3)
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1/4 = 0.25
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1/5 = 0.2
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1/6 = 0.1(6)
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1/7 = 0.(142857)
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1/8 = 0.125
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1/9 = 0.(1)
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1/10 = 0.1
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Where 0.1(6) means 0.166666..., and has a 1-digit recurring cycle. It can be
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seen that 1/7 has a 6-digit recurring cycle.
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Find the value of d < 1000 for which 1/d contains the longest recurring cycle
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in its decimal fraction part.
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"""
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def solution(numerator: int = 1, digit: int = 1000) -> int:
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"""
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Considering any range can be provided,
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because as per the problem, the digit d < 1000
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>>> solution(1, 10)
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7
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>>> solution(10, 100)
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97
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>>> solution(10, 1000)
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983
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"""
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the_digit = 1
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longest_list_length = 0
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for divide_by_number in range(numerator, digit + 1):
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has_been_divided = []
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now_divide = numerator
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for division_cycle in range(1, digit + 1):
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if now_divide in has_been_divided:
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if longest_list_length < len(has_been_divided):
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longest_list_length = len(has_been_divided)
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the_digit = divide_by_number
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else:
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has_been_divided.append(now_divide)
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now_divide = now_divide * 10 % divide_by_number
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return the_digit
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# Tests
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if __name__ == "__main__":
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import doctest
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doctest.testmod()
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