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The Goldstone Report will now go to the U.N.Security Council which could refer it to the International Criminal Court in The Hague.
VOA: standard.2009.10.18
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So, if we think about the second case here where we have c n minus, now we're talking about a molecule with a net charge of negative 1.
那么，如果我们考虑的是第二个例子,也就是氰离子，那么现在我们讨论的是一个净电荷量为负一的分子。
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Right now, I'm doing rock n' roll music.
现在我玩摇滚。
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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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In contrast, if we're taking the wave function and describing it in terms of n, l, m sub l, and now also, the spin, what are we describing here?
相反，如果我们考虑一个波函数,然后用ｎ，ｌ，ｍ小标ｌ,还有自旋，我们描述的是什么？
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But now executive secretary of the U.N.Framework Convention on Climate Change Yvo de Boer says any decision taken next month in Copenhagen will instead be "morally binding."
VOA: standard.2009.11.06
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So in other words, every time I merge the point that I kept emphasizing verbally there and that I'm only touching each number once, means that we have to account for the amount of time it takes to merge N which is going to be just N. Now, this is again one of these cyclical answers.
换言之，之前在做合并时,我不停地强调,对每个数字我只碰了一次，这就是说,我们要记录合并所花的时间量,也就是这里的,这又是一种循环性的答案。
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All U.N.personnel have now been barred from the entering the town.
VOA: standard.2009.08.27
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Such where the deep transported mind may soar Above the wheeling poles, and at Heav'n's door Look in, and see each blissful Deity Now the graver subject that Milton is intending at some point to expound upon is clearly an epic one.
埋藏于心底的思想会,冲天而上并到达上帝的门前,往里一瞥你会看到每个快乐的神,现在弥尔顿正打算用更庄重的主题,详细点说就是史诗。
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His followers are now targets of the Ugandan,south Sudanese, CAR,and the U.N.-backed Congolese forces, leading many to question what objective his latest moves could possibly be designed to reach.
VOA: standard.2009.08.27
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So, now we have n = 3 and in the chart are stocks, bonds, and oil.
图表里n=3表示,股票，债券和石油这三类资产。
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e The charge on the anion times minus e, so there is the minus e squared, 0R0 and divided by 4 pi epsilon zero r naught, because now I am evaluating this function at r naught, one minus one over n where n is the Born exponent.
阴离子的电荷乘以，因此会有-e的频繁，除以4πε,因为现在我用r圈评估这个函数，1-1/n，n是波恩指数。
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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。
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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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J++ Now here's the semicolon, J less than N, where N is this, J plus, plus; so what am I doing?
现在这里是一个分号，J小于N，N是这个，那么我在做什么？
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So instead of being equal to negative z squared, now we're equal to negative z effective squared times r h all over n squared.
这里不再等于-z的平方，现在我们等于-有效的z的平方,乘以RH除以n的平方。
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When we talk about the n equals 2 state, we now have 2 squared or 4 degenerate same energy orbitals, and those are the 2 s orbital.
当考虑n等于2的态时，我们有2的平方,或者4个简并能量的轨道，它们是2s轨道。
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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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Now, does this jive with our little asymptotic claim here, our little analysis with N notation?
那么这与用N表示法所，得出的结论是否一致呢？
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So, let's try another example here, and let's try a case now where instead of dealing with a neutral molecule we have an ion, so we have c n minus.
那么，让我们来试一下另外一个例子，这次不是一个中性原子,而是一个离子，氰离子。
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And this is now consistent with my claim that I have sorted a list of size N equals 1.
这与我之前所说的是一致的,我已经将N为1的一个序列排好了序。
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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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So now we have that energy is equal to the negative of the Rydberg constant divided by n squared.
我们可以把能量方程大大简化,现在能量等于负的Rydberg常数除以n平方。
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Now at this point in the story, I've got a variable called n. I've stored it in int.
现在这个故事的这里，我获得了一个变量n，我把它存储成int型的。
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I now have a list of size 1 so N is, in fact, less than 2.
现在序列的大小是1，可见N小于。
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If I'm using algorithm that I'm now calling merge sort, T the running time involved in sorting N elements, T of N, you know, is just the same as running the algorithm for the right half, plus what's this plus N come from?
如果我用归并排序算法，对N个元素其运行时间，就等于此算法一半元素的运行时间,另一半的运行时间，再加上N，这个N是什么呢？
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If I want to move this stack here, I'm going to take a stack of size n minus 1 move it to the spare spot, now I can move the base disk over, and then I'm going to move that stack of size n minus 1 to there. That's literally what I did, OK?
如果我想移动这些圆盘，我先把从n-1个圆盘1,移动到多余的柱子上去，这样我就能把最底下的圆盘放到这儿了，然后再把从n-1个圆盘放到这儿来,这就是确切的我怎么做的对不对？
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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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And now I have a problem of size N minus 1.
现在是一个N-1大小的问题。
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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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