For our step two, what we need is number of valence electrons.

我们的第二步，需要知道价电子的个数。

麻省理工公开课 - 化学原理课程节选

Count the number of primitive operations in each step.

数一数每一步中的基本操作,好的，如果我们看看这段代码。

麻省理工公开课 - 计算机科学及编程导论课程节选

So this line or these lines of code up here are arguably constant time steps to say if N is less than 2 in return, that it will always take maybe one step, maybe two steps, some number of fixed CPU cycles.

如果N小于2并返回,那么这些行所对应的代码,通常只需要执行一步，或者两步，具体数字与CPU周期有关。

哈佛公开课 - 计算机科学课程节选