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OK. So, soon as I get down to a list that has no more than two elements in it, I'm done.
不超过2个元素的列表，那就结束了,注意，这里是小于等于。
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So we have--U.S. Government issues bills and that's less than or equal to one year and they pay no interest.
美国政府发行的短期国库券,期限通常小于或等于一年,且不付息
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n Alright. So if n is less than or equal to 1, return n. Well that's not right, right?
好吧，如果n小于等于1，它会返回，这里不对，是吧？
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We discovered that the quantity dA, under conditions of constant volume and temperature, dA TS And A is u minus TS.
我们发现在恒定的体积和温度下，亥姆赫兹自由能的变化，小于零，is，less，than，zero。,亥姆赫兹自由能A等于内能u减去。
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00 Now I say while i is less than or equal to 100, I write my code, I sleep for a second.
现在我指明当i小于等于，我写下我的代码，我休眠一分钟。
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OK. If last minus first is greater than or equal to 2-- sorry, less than 2, then either look at this thing or look at that thing.
大于等于儿，抱歉，是小于二,看看这里或者那里,这就是我说的要特别小心的地方。
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It says, well I'm going to print out first and last just so you can see it, and then I say, gee 2 if last minus first is less than 2, that is, if there's no more than two elements left in the list, then I can just check those two elements and return the answer.
然后它计算了尾点和开始点的差,如果小于2的话,也就是说数组中的元素小于等于，我对这两个元素进行比较,然后返回结果就可以了，否则的话，我们就去寻找中值点，注意它是怎么实现的,首先这个指向一个列表的开头。
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I want less than, rather than less than or equal to.
对不对？我不应该写小于等于而应该写小于。
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We call an instrument of the U.S. Treasury with a maturity less than or equal to one year--we call that a bill and they used to be the only discount bonds issued by the U.S. Government.
我们把期限小于或等于一年的,国库券品种称为短期国库券,它们曾是美国政府,唯一发行的贴现债券
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I assert the counter is less than or equal to 0.
是小于等于0的。
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I can either look at my flow chart, or I can look at the code. If I look at the flow chart, it says, I'm at this point. Look at ANS squared. Is it less than or equal to-- sorry, first of all, 0 ANS squared is 0, is it less than or equal to x, yes.
我可以看流程图也可以看我的代码,如果我看流程图的话，流程图这么说的，在这一点，看看ANS的平方,看是不是小于等于-对不起，首先，ANS的平方是。
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So if n is greater than or equal to 1 and n is less than or equal to 3, let's just judge this thing a small number arbitrarily.
所以n大于等于1，并且n小于等于,我们武断的判定它为,一个小的数字。
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Going around in a cycle the integral of dq over T is less than or equal to zero.
对一个循环过程作dq除以T的积分,小于等于零。
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So what do I do? Change ANS. X doesn't change.
那么它是不是小于等于x呢？
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If it's not greater than x, take 2. Square it.
求它的平方，如果1的平方小于等于x的话。
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This checks "Is i less than or equal to one hundred?
这个检查“I是不是小于等于100？
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I test an end test. So the flow chart says, and the tradition was to do this in a diamond shape, I'm going to check if ANS times ANS-- oh, which way did I want to do this x - is less than or equal to x. Now that's a test.
首先进行终结测试，因此流程图就的去,传统的做法是把这个放到一个菱形里面,检查下是否ANS乘以ANS-哦,我刚才是怎么说的来着-是不是小于等于。
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-- Else if n is greater than 4 and n is less than 6 -- 7 let's call it medium -- else if n is greater than or equal to 7, -- less than or equal to 10 -- let's call it "big" -- and if the user typed in zero or negative 10 or 20 or whatever, let's just use the all-inclusive else block and just say, "You picked an invalid number."
否则如果n大于4，并且n小于-,我们叫它为中间数--否则如果n大于等于，小于等于10--我们叫它为“大数“,如果用户键入了0或者-10或20或其它的，让我们用一个广泛的else子句，只需要说，“你选了一个非法的数字“
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