n l m s Once we have chosen a certain mix of n, l, m and s, it is used once for that particular atom.

一旦我们选定了一组量子数，它就只能被一个固定原子所有。

麻省理工公开课 - 固态化学导论课程节选

So in other words, every time I merge the point that I kept emphasizing verbally there and that I'm only touching each number once, means that we have to account for the amount of time it takes to merge N which is going to be just N. Now, this is again one of these cyclical answers.

换言之，之前在做合并时,我不停地强调,对每个数字我只碰了一次，这就是说,我们要记录合并所花的时间量,也就是这里的,这又是一种循环性的答案。

哈佛公开课 - 计算机科学课程节选

But, once you get beyond n equals one, 2s2 2p6 3s2 3s6 it's always 2s2, 2p6, 3s2, 3s6.

但是，一旦你让n大于，它就总是。

麻省理工公开课 - 固态化学导论课程节选

Once I have it sorted I can search it in log n time, but that's still isn't as good as just doing n. And this led to this idea of amortization, which is I need to not only factor in the cost, but how am I going to use it?

一旦对其完成排序,就可以在log，n的时间内对其完成搜索,但是这样做仍然不如n的复杂度,这样做引出了耗时分摊的想法,这时不仅需要考虑耗时的因素？

麻省理工公开课 - 计算机科学及编程导论课程节选