50 lines
836 B
Python
50 lines
836 B
Python
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# Problem 1:
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#
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# If we list all the natural numbers below 10 that
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# are multiples of 3 or 5, we get 3, 5, 6 and 9.
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#
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# The sum of these multiples is 23.
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#
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# Find the sum of all the multiples of 3 or 5 below 1000.
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def f_timer(passed_f):
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def inner_f(*args, **kwargs):
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import time
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t1 = time.time()
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value = passed_f(*args, **kwargs)
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t2 = time.time()
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t3 = t2 - t1
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print("The function {} ran for {:4f} seconds... ".format(passed_f.__name__,t3))
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return value
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return inner_f
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def check_divisibility(num, dict_multiples: dict) -> bool:
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dummy = 1
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for m in dict_multiples:
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dummy *= num%m
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if dummy == 0:
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return True
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else:
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return False
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@f_timer
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def main():
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multiples = {3,5}
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sumval = 0
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for n in range(1000):
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r = check_divisibility(n,multiples)
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if r:
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sumval += n
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print(sumval)
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main() |