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And he said the number of facilities supplying anti-retroviral drugs used to combat AIDS would be doubled from 500 to 1,000.
VOA: standard.2010.04.26
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And we can calculate that with the formula that we used, which was just n minus l minus 1 equals the number of nodes.
这个我们可以用我们以前用过的那个公式来计算，也就是节点数等于ｎ减去ｌ减去１
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All right, so the bonding order, you're correct, should be 2, if we subtract the number of bonding minus anti-bonding electrons and take that in 1/2.
好，你们是对的，键序为，如果我们用成键数,减去反键数除以2。
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Under the plan, the government would also triple the number of condoms distributed each year to 1.5 billion.
VOA: standard.2010.04.26
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I'm then printing bracket one, bracket two, bracket three, bracket dot, dot, dot up until the total number of arguments, 1 whatever that may be, and it's going to be at least one because the program always has a name.
然后我打印，括号等等等，知道打印完所有的参数，不管那个最大值可能是什么，它将至少为,因为程序总是有一个名字。
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Actually it's not 40, I think these are, 75 alright. So the answer is 75 and the number of calls is 1.7 million.
实际上不是4，我觉得，好了，所以答案是5,调用次数是1百70万次。
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Don't all come crashing into 1:00 because we don't have the room or the number of exams.
别再1：00的时候冲进来0,因为我们没有房间，或者考场号。
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How many of you, excluding the people who chose 1 last time, how many of you chose a number that was lower than the number you chose last time?
出了上次选1的人,多少人选了比你,上次选择的数还低的数字
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But from these two definitions appearing in book III chapter 6 and Book IV chapter 1 we learn a number of important things.
但从这两个出现在,第，III，册第，6，章和第，IV，册第，1，章的定义1,我们得知一些重要的事。
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The way that we can figure this out is using something called bond order, and bond order is equal to 1/2 times the number of bonding electrons, minus the number of anti-bonding electrons.
我们可以用叫做,键序的概念来弄明白它，键序等于1/2乘以成键电子,数目减去反键电子数目。
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So what we end up with for the number of photons in this laser beam of light is 2.1 times 10 to the 17 photons.
这个激光束的数目是,2，1乘以10的17次方。
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So then in order to figure out the complete number of valence electrons in our molecule, we just add 5 plus 4 plus 1.
那么接下来为了得到,这个分子中价电子的总个数，我们只需要将五加上四，再加上一。
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I've included item number 1, which has a value of 7. Right?
我已经装了1号物品，1号物品的值是7，对不对？
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I bet there are people in this room who know the lyrics to over 1,000 songs, NBA know the number of every individual in the National Basketball Association, so I think that learning some subset of elements is not unreasonable.
我敢打赌这个房间里有那么些人,他们能记得住超过1000首歌的歌词，记得住,每个人的号码，因此我认为,能记住这些元素也不是不可能的。
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We let the number of pairs at month 1 be 1.
实际上不是应该是1。
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So for the bond order we want to take 1/2 of the total number of bonding electrons, so that's going to be 4 minus anti-bonding is 4, so we end up getting a bond order that's equal to 0.
键序等于1/2乘以,总的成键电子数，也就是4，减去反键电子数，也就是4。,所以最后得到键序为0。
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You can actually say big O of 1, big O of 1 being constant time, the same number of steps.
其复杂度为O，表示时间是一个常量，所用的步数是相同的。
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The question was how many rabbits do you have at the end of a year? At the end of two years? At the end of more than that number of years, and so. We can do this with the following level definition. We're going to let pairs of 0, the number of pairs at month 0, actually it would not be it would be 1.
在下一个月末的时候让我们假设,这2只小兔子也有了后代,同样是一公一母,问题是一年后你会有多少只小兔子？,两年后呢？更多年后呢？,我们可以用接下来的等级说明来解释，在第0个月有0对兔子。
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We give you a very rich table of constants that's got all kinds of things from the mass of the electron to the speed of light, and all this stuff to the requisite number of significant figures. And, in addition, you are allowed to take in one sheet of paper, 8 1/2 x 11 one sheet 8 1/2 x 11, you can write anything you want on it.
我们会给你一个很详实的常量表,将会涉及很多方面，从元素的电子,到光速，这些内容到有效数字的定量,还有，你们可以带来一张纸，纸的规格是，可以写任何你们想写的东西。
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And then the number of pairs at month n is the number of pairs at month n - 1 plus the number of pairs at month n - 2.
我们让第一个月的兔子数是1对,第n个月的兔子对数,是第n-1个月的兔子对数。
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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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And then 1/2 of the number of shared electrons.
然后再减去共用电子的个数的一半。
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I've got one test, I've got a subtraction, I've got a multiplication, that's three steps, plus whatever number of steps it takes to solve a problem of size b minus 1.
我进行了一次比较，一次减法，一次乘法，一共是三个步骤,再加上t的步骤数。
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