Question
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.
思路
- 暴力列举所有包含3，5因子的数，进行求和
- 1000以内以3，5为因子的数的和，即考虑能被3，5各自整除的数的和，再去除能被15整除的数即可。
Python 代码
暴力列举法
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def (n):
all_sum = 0
for i in range(1, n):
if i%3 == 0:
all_sum += i
elif i%5 == 0:
all_sum += i
return all_sum
sum_of_multiples1(1000)
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233168
公式法
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def sum_of_multiples(n):
third_count = (n-1) // 3
five_count = (n-1) // 5
fifteen_count = (n-1) // 15
print(third_count, five_count, fifteen_count)
all_sum = 3 * (third_count)*(1+third_count)/2 + 5 * (five_count)*(1+five_count)/2
- 15 * (fifteen_count)*(1+fifteen_count)/2
return int(all_sum)
sum_of_multiples(1000)
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333 199 66
233168
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