His adventurous song is one that "with no middle flight intends to soar above th' Aonian Mount, while it pursues things unattempted yet in Prose or Rhyme."

他那具有冒险精神的诗歌,“抛开了中途的旅程，要一飞冲天越过爱奥尼神山，同时追寻使用在未经使用的韵律“

耶鲁公开课 - 弥尔顿课程节选

This is the initial part of the simile that describes the leaves: "thick as Autumnal Leaves that strow the Brooks / in Vallombrosa, where th' Etrurian shades / high overarch't imbow'r."

这是描写落叶的比喻的开始部分：,“他们人数众多，稠密得像秋天的繁叶,/在瓦隆布罗萨，厄特鲁利亚的光辉逐渐暗淡“

耶鲁公开课 - 弥尔顿课程节选

Just swapping them, right? I temporarily hold on to what's in the i'th element so I can move the i plus first one in, and then replace that with the i'th element.

交换他们，对么？,临时的保存下第i个元素,然后把第i+1个元素移进来，把i+1的位置替换为第i个元素。

麻省理工公开课 - 计算机科学及编程导论课程节选

I've got to count my way down, which means that the access would be linear in the length of the list to find the i'th element of the list, and that's going to increase the complexity.

的位置并去访问，然后继续下去，也就意味着,找到数组中的第i个元素的方法,是关于数组的长度呈线性复杂度的，这回增加算法的复杂度。

麻省理工公开课 - 计算机科学及编程导论课程节选

With this, if I can assume that accessing the i'th element of a list is constant, then you can't see that the rest of that analysis looks just like the log analysis I did before, and each step, no matter which branch I'm taking, I'm cutting the problem down in half.

读取数组中的第i个元素,是个常量时间的操作的话，我也就能像以前那样得到,这个算法是对数级复杂度的分析,并且每一步不管我选择哪个区间，我都可以把问题的规模缩小一半。

麻省理工公开课 - 计算机科学及编程导论课程节选