110 lines
2.9 KiB
Python
110 lines
2.9 KiB
Python
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from copy import deepcopy
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from itertools import islice
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def primes(n):
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""" Nice way to calculate primes. Should be fast. """
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l = range(2, n + 1)
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_l = []
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while True:
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p = l[0]
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if p * p > n:
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return _l + l
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l = [i for i in l if i % p != 0]
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_l.append(p)
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def calculate_decimal_places(numerator, denominator):
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numerator = (numerator - (numerator / denominator) * denominator) * 10
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digits = []
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while True:
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digit = numerator / denominator
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numerator = (numerator - digit * denominator) * 10
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digits.append(digit)
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if digits[-3:] == [0, 0, 0]:
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raise StopIteration
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yield digit
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def has_cycle(decimal_places, n):
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d = decimal_places
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return list(islice(d, 0, n)) == list(islice(d, 0, n)) and \
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list(islice(d, 0, n)) == list(islice(d, 0, n))
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def f_025_1():
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""" Second try. I realized that only primes must be
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checked. Therefore, my brute force approach worked. """
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l = []
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for d in primes(1000):
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for i in range(5, 10000):
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decimal_places = calculate_decimal_places(1, d)
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if has_cycle(decimal_places, i):
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l.append((i, d))
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break
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print(max(l))
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def calculate_cycle(numerator, denominator):
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numerator = (numerator - (numerator / denominator) * denominator) * 10
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remainders = set([])
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while True:
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digit = numerator / denominator
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remainder = (numerator - digit * denominator)
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if remainder in remainders:
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raise StopIteration
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remainders.add(remainder)
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numerator = remainder * 10
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yield digit
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def f_025_2():
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""" Understood trick with remembering remainder... """
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s = [(len(list(calculate_cycle(1, d))), d)
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for d in range(1, 1001)]
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print(max(s))
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def f_025_3():
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""" Only testing primes... """
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s = [(len(list(calculate_cycle(1, d))), d)
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for d in primes(10000)]
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print(max(s))
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f_025_3()
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#print([(find_cycle_count(calculate_decimal_places(1, d)), d)
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# for d in range(1, 100)])
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#print(find_cycle_count(calculate_decimal_places(22, 7)))
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#l = []
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#for d in range(1, 1000):
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# for i in range(5, 1000):
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# decimal_places = calculate_decimal_places(1, d)
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# if has_cycle_one_off(decimal_places, i):
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# l.append((i, d))
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# break
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# decimal_places = calculate_decimal_places(1, d)
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# if has_cycle(decimal_places, i):
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# l.append((i, d))
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# break
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#def find_cycle_count(decimal_places):
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# cycles = []
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# for digit in decimal_places:
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# new_cycles = []
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# for cycle in cycles:
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# digits, length = cycle
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# if digits[0] == digit:
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# if len(digits[1:]) == 0:
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# return length
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# new_cycles.append((digits[1:], length))
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# new_cycles.append((digits + [digit], length + 1))
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# new_cycles.append(([digit], 1))
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# cycles = new_cycles
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