Let's say you have to go through three or four operations to get a final number, well, do it algebraically.

让我们说,你不得不通过3-4个操作,才能得到最终的数，好吧，用代数方法求解。

麻省理工公开课 - 固态化学导论课程节选

Up here I have n over two operations of size two. Up here I've got n over four operations of size four.

最下面有n个规模为1的操作,接着上面有n/2个规模为2的操作,再上面有n/4个规模为4的操作。

麻省理工公开课 - 计算机科学及编程导论课程节选

Down here, I've just got two things to merge, and then I've got things of size two to merge and then things of size four to merge. But notice a trade off. I have n operations if you like down there of size one.

但是n的大小是不同的，是吗？在这里我们只要合并两个元素,然后是合并长度为2的列表,接下来是合并长度为4的列表,但是观察一下之间的权衡关系。

麻省理工公开课 - 计算机科学及编程导论课程节选