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Last time we checked the type and said if it is a float you're okay. If not, carry on.
并且会告之是否是一个浮点数,如果不是继续循环。
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It has to have an origin of replication, which we talked about last time, which is compatible with the cells.
所以质粒必须带有复制起点,我们上次讲过,该起点要与受体细胞匹配
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That's certainly one way to do it, but for computers that's not necessarily the easiest way. So another way of solving it is to do something we already saw last time, which is basically, why not simply enumerate all possible examples and check them?
将一个等式代入另一个等式，这当然是一种办法,但是对电脑来说这绝对不是一件简单的事,所以解决这个问题的办法，正如我们上一次看过的,非常基本？
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Taking the problem, recognizing that you know what, 8 even though this is a pretty big problem size 8 in this case and last time it was size 8 or in the case of the papers in size of a thousand roughly with the phonebook, I assume these are in a perfectly straight line they won't quite fit.
以这个问题为例，你们要认识到，在这种情况下，这是个比较大的问题，其大小是,上次它的大小也是8，但在纸片那个问题中,电话簿的规模大概是上千的，现在假设这些,杯子完全在同一条直线上，虽然并不十分符合这个条件。
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If there was a time to go too fast it was probably last time, because it turns out this is something that's covered well in the textbook, so it's covered in chapter six of the textbook.
如果真有哪次讲得太快,可能就是上次,因为这些内容,已经在课本上写得很清楚了,就在课本的第六章
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he is not here today the person held it ? oh, say it again the person bring it in last time He's not here today -Oh ,he's not here today...yeah ? can any of you manage this?
他今天没来,上次拿碟的人,啊？什么,上次拿了碟的人,他今天不在,你们谁拿一下这个？
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All right, so one thing that I want to point out, which I said many, many times on Friday, and this is perhaps the last time I'll say it, but one last time is we can think about why we only see a line for the 2 p orbital, versus we don't see separate lines for a 2 p x, a 2 p y, and a 2 p z.
好的，我还要指出一个问题，这个问题我在上周五已经说了很多很多次了，这可能是我最后一次提到它，但是这最后一次让我们来考虑一下,为什么我们只看到了一条，对应于，2，p，轨道的线,而不是分别对应于，2，p，x，2，p，y，2，p，z，的线？
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to the n, every value in the 1 bit vector we looked at last time is either 0 or 1. So it's a binary n number of n bits, 2 to the n.
从2到n，我们上次看到的,位向量的每个值不是0就是,所以它是n，比特的二进制数,从2到。
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