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24fb140672
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233fc85708
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| 233fc85708 | |||
| 2fb728a4ba | |||
| e900cb117e | |||
| a036c30fc4 | |||
| 8c3da33225 |
@@ -22,7 +22,7 @@ def create_problem():
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https://projecteuler.net/problem={args["problem"]}
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https://projecteuler.net/problem={args["problem"]}
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"""
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"""
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from utils import timeit
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from project_euler_python.utils import timeit
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@timeit("Problem {(args["problem"]):0>3}")
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@timeit("Problem {(args["problem"]):0>3}")
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@@ -34,7 +34,7 @@ def compute():
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names = sorted(f.read().replace('"', "").split(","))
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names = sorted(f.read().replace('"', "").split(","))
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ans = 0
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ans = 0
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for idx, name in enumerate(names, 1):
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for idx, name in enumerate(names, start=1):
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ans += sum(ord(char) - 64 for char in name) * idx
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ans += sum(ord(char) - 64 for char in name) * idx
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return ans
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return ans
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@@ -9,9 +9,21 @@ Solution for problem 023 of Project Euler
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https://projecteuler.net/problem=23
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https://projecteuler.net/problem=23
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"""
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"""
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from itertools import combinations_with_replacement
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from project_euler_python.utils import timeit
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from project_euler_python.utils import timeit
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def _get_abundant_numbers(limit):
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divisor_sums = [0 for _ in range(limit)]
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for divisor in range(1, limit):
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for multiple in range(divisor * 2, limit, divisor):
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divisor_sums[multiple] += divisor
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return [number for number in range(1, limit) if divisor_sums[number] > number]
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@timeit("Problem 023")
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@timeit("Problem 023")
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def compute():
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def compute():
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"""
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"""
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@@ -35,25 +47,19 @@ def compute():
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sum of two abundant numbers.
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sum of two abundant numbers.
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"""
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"""
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limit = 28124
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limit = 28123 + 1
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divisor_sum = [0] * limit
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abundant_numbers = _get_abundant_numbers(limit)
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for i in range(1, limit):
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for j in range(i * 2, limit, i):
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divisor_sum[j] += i
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abundant_nums = [i for (i, x) in enumerate(divisor_sum) if x > i]
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expressible_sums = {
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first + second
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for first, second in combinations_with_replacement(
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abundant_numbers,
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2,
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)
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if first + second < limit
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}
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expressible = [False] * limit
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return sum(number for number in range(1, limit) if number not in expressible_sums)
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for i in abundant_nums:
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for j in abundant_nums:
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if i + j < limit:
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expressible[i + j] = True
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else:
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break
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ans = sum(i for (i, x) in enumerate(expressible) if not x)
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return ans
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if __name__ == "__main__":
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if __name__ == "__main__":
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@@ -9,7 +9,7 @@ Solution for problem 024 of Project Euler
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https://projecteuler.net/problem=24
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https://projecteuler.net/problem=24
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"""
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"""
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from itertools import permutations
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from itertools import islice, permutations
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from project_euler_python.utils import timeit
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from project_euler_python.utils import timeit
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@@ -28,9 +28,10 @@ def compute():
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0, 1, 2, 3, 4, 5, 6, 7, 8 and 9?
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0, 1, 2, 3, 4, 5, 6, 7, 8 and 9?
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"""
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"""
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limit = 1_000_000
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digits = list(range(10))
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digits = list(range(10))
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_permutations = list(permutations(digits))
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millionth_permutation = next(islice(permutations(digits), limit - 1, limit))
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ans = "".join(str(digit) for digit in _permutations[999_999])
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ans = "".join(str(digit) for digit in millionth_permutation)
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return ans
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return ans
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@@ -37,11 +37,17 @@ def compute():
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cycle_length = 0
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cycle_length = 0
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ans = 0
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ans = 0
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# Because denominators with factor 2 contribute only to the terminating
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# part of a decimal. The repeating behavior comes from the part of the
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# denominator that is coprime with 10, so evens are skipped
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for number in range(3, 1000, 2):
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for number in range(3, 1000, 2):
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if number % 5 == 0:
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if number % 5 == 0:
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continue
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continue
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p = 1
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p = 1
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while 10**p % number != 1:
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remainder = 10 % number
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while remainder != 1:
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# avoid huge exponentiation
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remainder = (remainder * 10) % number
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p += 1
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p += 1
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if p > cycle_length:
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if p > cycle_length:
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cycle_length, ans = p, number
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cycle_length, ans = p, number
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