Best Python code snippet using assertpy_python
fizzbuzz.py
Source:fizzbuzz.py
1import sys2def is_multiple_of(a: int, b: int):3 """Check if a is a multiple of b"""4 return a % b == 05def fizzbuzz(array_size, fizz, buzz):6 """7 Print 'Fizz' if multiple of fizz variable, 'Buzz' if multiple of8 buzz variable and 'FizzBuzz' if multiple of both.9 """10 for i in range(1, array_size+1):11 value_to_print = i12 if is_multiple_of(i, fizz) and is_multiple_of(i, buzz):13 value_to_print = "FizzBuzz"14 elif is_multiple_of(i, fizz):15 value_to_print = "Fizz"16 elif is_multiple_of(i, buzz):17 value_to_print = "Buzz"18 print(value_to_print)19if __name__ == '__main__':20 arguments = sys.argv[1:]21 try:22 parameters = [int(x) for x in arguments]23 except ValueError:24 raise ValueError("Apenas valores inteiros são aceitos.")25 integers_array_size = parameters[0]26 fizz = parameters[1]27 buzz = parameters[2]28 if integers_array_size < 2:29 raise ValueError("O tamanho da lista deve ser maior que 1.")30 fizzbuzz(integers_array_size, fizz, buzz)
main.py
Source:main.py
...3If we list all the natural numbers below 10 that are multiples of 3 or 5, we get 3, 5, 6 and 9. The sum of these multiples is 23.4Find the sum of all the multiples of 3 or 5 below 1000.5"""6N = 10007def is_multiple_of(left, right):8 return right % left == 0 or left % right == 09if __name__ == "__main__":10 summed = 011 for i in range(2, N):12 if is_multiple_of(i, 3) or is_multiple_of(i, 5):13 summed += i...
sum_of_multiples.py
Source:sum_of_multiples.py
1def sum_of_multiples(limit, multiples):2 multiples = [n for n in range(1, limit) if is_multiple_of(n, multiples)]3 return sum(multiples)4def is_multiple_of(n, multiples):5 for m in multiples:6 if n % m == 0:7 return True...
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