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So this line or these lines of code up here are arguably constant time steps to say if N is less than 2 in return, that it will always take maybe one step, maybe two steps, some number of fixed CPU cycles.
如果N小于2并返回,那么这些行所对应的代码,通常只需要执行一步，或者两步，具体数字与CPU周期有关。
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So, step five tells us to add 2 electrons between each atom, so we add two there.
那么，第五步告诉我们在两个原子之间放上两个电子，因此我们在这放上两个。
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And I'm going to show you an example in a 1 second, just to drive this home, but notice the characteristics. In the first two cases, the problem reduced by 1 at each step.
在前面两个例子里,每一部问题的规模缩小了,不管是迭代的还是递归的，这表明这个问题的复杂性可能是线性的。
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Probably I shouldn't have said it's a two-step process 'cause it's a three-step process to actually write and run it, and the third step is just going to be to run it.
也许我不应该说它是个两步完成的程序,准确来讲，应该是三步，前两步--写，第三步--运行。
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But what happens is you can do a two-step process.
但你可以采取两步法。
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Let's take a step back and ask are all kinds of cells really the same and start by talking about two cells that I know you know are quite different.
让我们退一步并问一句,所有细胞真的都是一样的么,我们不妨从两种,看起来截然不同的细胞谈起
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What we've done is taken this pitch and played it all the way down an octave below it so we're actually getting back to this configuration of the pitch right next to it, and we could--then of course we could go down one more step and we would get the octave, which is a duplication of two-to-one.
我们所做的是固定一个音高，向下一路弹奏一个八度,所以事实上我们又回到了与这个固定音高相邻的音上，我们当然也可以向下再弹一组，然后得到这个八度，这其实就是两个八度对一个八度的重复。
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So now, then we're going to use what we're going to learn from step two in order to calculate III this part, what we could call step three.
现在我们要用从步骤II中,得知的东西来计算这个部分,我们可以叫它步骤。
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So writing a program at least in this language is a two-step process.
用这种语言写一个至少需要,两步完成的小程序。
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For our step two, what we need is number of valence electrons.
我们的第二步，需要知道价电子的个数。
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So it turns out that there's a two-step process.
好像有两步要走。
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Going from two to three, that's an adiabatic expansion, so q is equal to zero in that step.
从第二点到第三点，是绝热膨胀，因此q等于零。
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