68 lines
1.8 KiB
Python
68 lines
1.8 KiB
Python
#!/usr/bin/python3
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# A unit fraction contains 1 in the numerator. The decimal representation of the unit fractions with denominators 2 to 10 are given:
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#
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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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#
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# Where 0.1(6) means 0.166666..., and has a 1-digit recurring cycle. It can be seen that 1/7 has a 6-digit recurring cycle.
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#
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# Find the value of d < 1000 for which 1/d contains the longest recurring cycle in its decimal fraction part.
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from timeit import default_timer
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def main():
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start = default_timer()
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max_ = 0
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for i in range(2, 1000):
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j = i
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# The repeating cycle of 1/(2^a*5^b*p^c*...) is equal to
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# that of 1/p^c*..., so factors 2 and 5 can be eliminated.
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while j % 2 == 0 and j > 1:
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j = j // 2
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while j % 5 == 0 and j > 1:
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j = j // 5
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# If the denominator had only factors 2 and 5, there is no
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# repeating cycle.
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if j == 1:
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n = 0
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else:
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n = 1
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k = 9
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div = j
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# After eliminating factors 2s and 5s, the length of the repeating cycle
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# of 1/d is the smallest n for which k=10^n-1/d is an integer. So we start
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# with k=9, then k=99, k=999 and so on until k is divisible by d.
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# The number of digits of k is the length of the repeating cycle.
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while k % div != 0:
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n = n + 1
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k = k * 10
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k = k + 9
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if n > max_:
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max_ = n
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max_n = i
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end = default_timer()
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print('Project Euler, Problem 26')
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print('Answer: {}'.format(max_n))
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print('Elapsed time: {:.9f} seconds'.format(end - start))
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if __name__ == '__main__':
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main()
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