Get busy, give charity. It's a problem that never goes away, so you can never rest.

忙碌劳作，给予施舍，这是一个永远不会消失的问题,所以你永远没办法停歇。

耶鲁公开课 - 旧约导论课程节选

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