您可以将循环扩展为以下内容:

You can expand the loops to something like this:

i = 1 ——>   1,2,3,…,b     b
i = 2 ——>   1,3,5,…,b     (b/2)
i = 3 ——>   1,4,7,…,b     (b/3)
i = 4 ——>   1,5,9,…,b     (b/4)
  …
i = b ——>   1, b          (b/b = 1)

这会扩展为以下形式的总和:

This expands into a sum of the form:

b + b/2 + b/3 + … + b/b = b * (1 + 1/2 + 1/3 + … + 1/b)

您可能认识到第二个因素是谐波系列.然后，使用以下SO答案的结果:找到谐波系列的Big O您可以获得嵌套循环的大哦":

You might recognize the second factor as the Harmonic Series. Then, using the result from the following SO answer: Finding Big O of the Harmonic Series you can get the Big Oh of your nested loops:

O(b * log(b))