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Creating rural jobs is on everyone's to-do list, from the government to the U-N to aid groups.
VOA: standard.2010.03.30
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What I am going to do now is I am going to multiply by N Avogadro and then add Born repulsion.
我接下来要做的是,将其乘以阿伏加德罗常数,再加入Born的排斥作用。
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n So I've got to do n things n times.
所以就是做n次。
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What does the U.N.plan to do with the data it has gathered now?
VOA: standard.2010.07.07
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For n = --now I'm going to do n = 100.
对于n=...我假设n=100
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So, the first thing we do is put two electrons between h and c, and then two electrons between c and n.
那么，我们先在氢和碳之间放两个电子，然后在碳和氮之间再放两个电子。
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Now, it's kind of a cyclical argument here because how do you sort the left half of N elements?
在这儿有点循环的意思,那么如何对N个元素中的左半部分进行排序呢？
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He constantly warns his son, who was kind of an n'er do well, that, "You better work hard, " Or, "You better work a little harder and pay more attention to what you're doing.
他不断提醒那一事无成的儿子,"你要加倍用功,哪怕只是稍微用点心,对你做的事要更上心
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I could do this. I could call this n equals one, the fundamental frequency.
我可以这样做，我将n等于，基频。
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Ah, n times, because for each value of i, I'm going to do that m thing, n*m so that is, close to what you said, right?
因此这就和你说的差不多了对不对？,这个问题的复杂度为，让我写下来，是-对不对，是？
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Given the code up there, if I want to move a tower of size n, what do I have to do?
鉴于这里的代码，如果我想移动N个圆盘，我该怎么做？
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So, let's, for example, look at nitrogen. So n 2, we can do the chart here in green, so it's the green dotted line, and what we see is that we have now defined this energy as where the 2 nitrogen atoms are separated.
那么，让我们举个例子，看一下氮，那么氮分子，我们可以把它用绿色曲线画在这，这是绿色的虚线，可以看到,我们已经定义为零点能，当两个氮原子分离时。
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At this point in the story, I now check while n is less than 1, what do I do?
在此时此刻，我现在在检查当n小于1是，我做什么？
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So, the optimal thing to do if you live in a world like this n is to get n as large possible and you can reduce the standard deviation of the portfolio very much and there's no cost in terms of expected return.
如果现实中也这样简单的话,那么你就尽量增大,这样就能让投资组合的标准差,就会大大降低,从预期收益率的角度来看,这样做的成本是零。
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If we can sort things, you know, we get this n log n behavior, and we got a n log n behavior overall. But can we actually do better in terms of searching.
如果我们可以排序，如你所知，我们有n，log，n级别的算法，并且我有一个整体的n，log，n级别的算法,但是我们在搜索方面可以做的更好吗？
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printf So thus far, the things I put David between quotes are just simple things like David or David backslash N, but what if I want to do call my self David in quotes, right?
像我们前面提到的,我们把要显示的东西放在双引号之间，譬如,或，David反斜杠n，但是如果我自己的名字本来就叫,“David“
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So instead you'd have to maybe if you start with wavelength, go over there, and then figure out velocity and do something more like kinetic energy equals 1/2 n b squared to get there.
这时你要先从波长开始，到这，然后算出速度,然后像动能等于1/2nb平方得到这。
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All right, I want to do one more at homonuclear example here, and this is n 2.
好的，我想要再讲一个同核的例子，这就是N2。
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But that merging process only takes N steps, N*log N so that's N times log of N. Now, it's a little tricky to reason through this perhaps the first time, let's just take a very simple example and see if we can do a little sanity check here.
但这个合并过程只需要N步，所以时间复杂度是,第一次对此进行推论可能会有点儿棘手，我们举一个简单的例子,看看我们能否做一些完整性的检查。
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So it feels a little backward mentally perhaps, 1 but this is saying, "do this block of code as long as n is less than 1."
所以它感觉可能有点迟缓，但是它指明，“只要n小于,执行这段代码“
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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。
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How do I rea-- replace the expression FOR t of n minus 1? Substitute it in again.
我们怎么来代替t这个表达式呢？,再来替换掉它，t等于。
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So what we'll do is this problem here, which is let's calculate out what the wavelength of radiation n would be emitted from a hydrogen atom if we start at the n equals 3 level and we go down to the n equals 2 level.
我们来做这个问题,让我们来计算一下,从n等于3到,等于2能级氢原子辐射的波长是多少。
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So we know that we can relate to z effective to the actual energy level of each of those orbitals, and we can do that using this equation here where it's negative z effective squared r h over n squared, we're going to see that again and again.
我们知道我们可以将有效电荷量与,每个轨道的实际能级联系起来，我们可以使用方程去解它,乘以RH除以n的平方,它等于负的有效电荷量的平方,我们将会一次又一次的看到它。
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If it doesn't say anything else we do mean n equals 1.
如果没有别的条件我们都指n等于。
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So this code is identical functionally nonswitch c to the last implementation we saw, nonswitch.c, but I'm just ever-more emphatically saying, "In case 1," that is when n equals 1 -- or when case 2 applies -- when n equals 2 or when n equals 3 do what?
所以这些代码到最后的实现上，功能是,完全相同的，但是我想要再次强调一下，“在case，1“，那是当n等于1时1,或者当case，2适用-,当n等于2或者当n等于3，它将做什么？
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And the last thing we can think about is how do we name this n h bond, and again, we just name it based on it symmetry.
最后我们要讨论的是,如何命名这个NH键，同样，我们基于它的对称性命名它。
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So if we go to the ground state, what you see is we're at that lowest energy level, and we only have one possibility for an orbital, because when n equals 1, that's all we can do.
如果我们在基态上，你可以看到，我们在能量最低的态上，只有一种，可能的轨道，因为n等于1，只有这种可能。
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So I always have to do a merge of n elements.
所以我始终需要做n个元素的合并。
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I could still do the linear case, which is order n or I could say, look, take the list, let's sort it and then search it. But in that case we said well to sort it was going to take n log n time, assuming I can do that.
我仍然可以做O的线性搜索,或者也可以以这个列表为例，我们先将其进行排序,然后再进行查找，但是在这种情况下,要花费n，log，n的时间去对其进行排序。
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