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68 lines
1.6 KiB
Python
68 lines
1.6 KiB
Python
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"""
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145 is a curious number, as 1! + 4! + 5! = 1 + 24 + 120 = 145.
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Find the sum of all numbers which are equal to the sum of the factorial of their digits.
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Note: As 1! = 1 and 2! = 2 are not sums they are not included.
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"""
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def factorial(n: int) -> int:
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"""Return the factorial of n.
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>>> factorial(5)
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120
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>>> factorial(1)
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1
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>>> factorial(0)
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1
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>>> factorial(-1)
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Traceback (most recent call last):
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...
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ValueError: n must be >= 0
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>>> factorial(1.1)
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Traceback (most recent call last):
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...
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ValueError: n must be exact integer
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"""
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if not n >= 0:
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raise ValueError("n must be >= 0")
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if int(n) != n:
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raise ValueError("n must be exact integer")
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if n + 1 == n: # catch a value like 1e300
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raise OverflowError("n too large")
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result = 1
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factor = 2
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while factor <= n:
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result *= factor
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factor += 1
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return result
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def sum_of_digit_factorial(n: int) -> int:
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"""
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Returns the sum of the digits in n
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>>> sum_of_digit_factorial(15)
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121
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>>> sum_of_digit_factorial(0)
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1
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"""
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return sum(factorial(int(digit)) for digit in str(n))
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def compute() -> int:
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"""
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Returns the sum of all numbers whose
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sum of the factorials of all digits
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add up to the number itself.
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>>> compute()
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40730
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"""
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return sum(
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num
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for num in range(3, 7 * factorial(9) + 1)
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if sum_of_digit_factorial(num) == num
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)
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if __name__ == "__main__":
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print(compute())
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