Problem 1

Problem 1

# Problem_1.py
"""
If 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.
Find the sum of all the multiples of 3 or 5 below 1000.
1000一下3的倍数以及5的倍数之和
"""

three = [i for i in range(0, 1000, 3)]
five = [i for i in range(0, 1000, 5)]
overlap = [i for i in three if i in five]

all = sum(three) + sum(five)
all -= sum(overlap)

print(all)