"That's kind of fun!" "What's your name?" "Broklin,B-R-O-K-L-I-N."

VOA: special.2010.04.30

The U.K. Had copied the U.S. Idea -I guess they copied -I don't know where the original -I think it's probably -is it U.K. Cpying the U.S. In this case?

英国效仿了美国的想法,我猜他们是效仿美国的,我也不知道谁先发明的,我认为可能是,存款保险是英国模仿美国

耶鲁公开课 - 金融市场课程节选

I mean, J. K. Rowling, Lloyd Alexander, C. S. Lewis, whatever you like.

比如说，罗琳、亚历山大、路易斯，你喜欢的都行。

英语阅读：技巧 - SpeakingMax英语口语达人

"But there were a couple "trouble songs" that just weren't working. They were o.k., but I wanted something different.

VOA: standard.2009.04.21

And if I pull it out one more level, 12k it's 12 plus 12 plus t of b over 8, 12 k because I'll have 12 of those to add up, plus t of b over 2 to the k.

总结一下也就是说，在k步以后，总步数应该是，那这种情况什么时候才能停止呢？,才能到达最基本的情况呢？

麻省理工公开课 - 计算机科学及编程导论课程节选

Bubba was my best good friend. I had to make sure he was O.K.

VOA: standard.other

If I had a third axis, I would draw a k, but we don't need that.

如果有第三根轴,我可以画一个矢量 k,但我们现在不需要它

耶鲁公开课 - 基础物理课程节选

O.K. But don't you be thinking that I'm goingto be calling you "sir."

VOA: standard.other

Why didn't they keep right on going with G, H, I, J, K, L, M, N, O, P?

为什么他们不接着写下去呢,按照字母表

耶鲁公开课 - 聆听音乐课程节选

O.K., follow me! Everywhere I went,I had to stand in line.

VOA: standard.other

The classic experiment that demonstrated anchoring was Kahneman and Tversky--I'm writing K&T -in a wheel of fortune experiment, which they did in 1974.

一个阐述锚定效应的经典实验是,卡尼曼和特维斯基...我用K&T来表示,他们在1974年做了幸运轮盘实验

耶鲁公开课 - 金融市场课程节选

O.K., Jenny, I'll stay. She was my most special friend.

VOA: standard.other

Yeah. You're jumping slightly ahead of me, but basically, I'm done when this is equal to 1 right? Because I get down to the base case, so I'm done when b u is over 2 to the k is equal to 1, and you're absolutely right, that's when k is log base 2 of b.

因为这就是最基本的情况了，因此当b/2的k次方等于1的时候就停止了1,你说的太对了，就是k等于b的以2为底的对数的时候,你们都坐的挺靠后的啊，我不知道是不是我讲的不太明白？

麻省理工公开课 - 计算机科学及编程导论课程节选

Still no shrimp, Lieutenant Dan. O.K., so I was wrong.

VOA: standard.other

If I reduce that it would be 3 plus t of b minus 3, so in general 3*k+t this is 3 k plus t of b minus k. OK.

把b减去一个,在外面加个3就可以了,因此也就是。

麻省理工公开课 - 计算机科学及编程导论课程节选

I'd always let her knowthat I was O. K.

VOA: standard.other

How long does it take me to find the k'th element? Linear. Because I've got to walk my way down it. OK? So in this case, you have linear access. Oh fudge knuckle.

线性的！因为我得从头,向下走一步步走,所以这里是线性访问，哦，有问题了吧。

麻省理工公开课 - 计算机科学及编程导论课程节选

No problem. Got to go.I'm double-parked. O.K. Thanks.

VOA: standard.other

Now I'm in better shape, right? Especially for really large n or for a lot of k, because now in general, this is going to be smaller than that.

很大的k也大的情况,因为这样这就比这个小了,这种情况是分摊。

麻省理工公开课 - 计算机科学及编程导论课程节选

O.K. Man, I tell you what.

VOA: standard.other

In the linear case, meaning in the unsorted case what's the complexity of this? k times n, right? Order n to do the search, and I've got to do it k times, so this would be k times n.

复杂度是多少？k的n次方，对吧？,在序列n中做搜索，要做k次，所以是k的n次方次,如果先排序后搜索。

麻省理工公开课 - 计算机科学及编程导论课程节选

Ugh. I'm O.K., Forrest.

VOA: standard.other

Yeah. Actually, I think I want b minus k equal to 1. Right?

对，实际上，当b-k=1的时候就该停下来了对不对？

麻省理工公开课 - 计算机科学及编程导论课程节选

I'm O.K. I'm fine.

VOA: standard.other

And what does that say? It says, let's assume I want to do k searches of a list. OK.

如果我们假定要在列表中做k次搜索,在线性的情况下，假定是一个未排序的情况。

麻省理工公开课 - 计算机科学及编程导论课程节选

Bubba. I'm O.K., Forrest.

VOA: standard.other

Here's the problem. How do I get to the nth- er, the k'th element in the list, in this case?

如何找到第n个元素呢-,在这里，如何找到第k个元素呢？

麻省理工公开课 - 计算机科学及编程导论课程节选

I'm O.K.

VOA: standard.other

So I'm done when b minus k equals 1, or k equals b minus 1.

因此b-k=1或k-b=1的时候，我就可以停下来了。

麻省理工公开课 - 计算机科学及编程导论课程节选