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I think it depends on whether the list is odd or even in length. Actually, that's probably not true. With one, it'll probably always get it down there, but if I've made it just equal to two I might have lost.
是奇数还是偶数，事实上,这是不正确的，如果最后剩下一个,那可能得到了结果,如果剩下两个，可能错了,所以，首先我们要格外。
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It's just like if you restrict me to only a subset of numbers, only the odd numbers, still there's an infinity of odd numbers.
这就像,如果你把我限制在数字的子集,只是奇数,但奇数的数量却仍然是无限多的
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Because it's now solving a one-smaller problem.
在上一步是奇数的基础上。
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Claire Elise is going to get her own vote plus 1, 2, 3, 4, 5, 6, 7, 8, 8 votes and Beatrice is getting 1, 2, 3, 4, 5, 6, 7, how did I end up with eight, I thought I had an odd number. 1, 2, 3, 4, 5, 6, 7 there must have been an odd number before.
克莱尔·伊莉斯将获得自己的选票加上,12345678 8张选票,而比阿特丽斯获得1234567,我怎么数出8个来的,我想应该是个奇数,1234567,之前也应该是个奇数
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In either case, I take that value and multiply back by two, if it was even I get back the original number, if it was odd, I'm not going to get back the original number, so I can just check to see if they're the same.
在两种情况中，我都把结果值,再乘以2，如果x是偶数我就,得到了原来的值，如果x是奇数那么就不会得到,原来的值，因此我看看得到的值,和原来的值相等不相等就可以了。
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It says, in either case in general, t of b-- and this is where I'm going to abuse notation a little bit but I can basically bound it by t, 12 steps plus t of b over 2.
我可以用一个,比12+t的数代表,这里有点不准确的地方,具体的步数依赖于奇数偶数，但是你们可以看到在两个case中。
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And in the even case, I'd square, divide by half, call this again: in the odd case, I go b minus 1 and then multiply by a.
就直接求a的平方的二分之b次方，如果是奇数的话，就去求a的b-1次方乘以a，以此递归下去。
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It's going to print out even or odd, depending on whether the value I put in, which is x in this case, is even or odd.
它会打印奇数或者偶数，这取决于我输入的值，也就是这个例子中的x，是奇数还是偶数。
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Remember last time, we wrote this simple piece of code to print out even or odd. If, you know, x, it was in fact, even or odd. So let me show you what a flow chart for that would look like, because I want to make a comparison point here.
记得上节课我们写过的,那个显示奇数还是偶数的程序么，如果，你知道，x，实际上是奇数还是偶数,那么让我来为大家画个这个程序的流程图吧，因为我这里想为大家做个比较。
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