设x为a加b之和。
B We just compute: X equals a inverse B.
我们只用这样算:X等于A逆乘。
What's the cheapest way to get X packages from point a to point B?
将X个包裹从a点送到b点的最便宜方法是什么?
Y He's not returning A or B or temp, and definitely not X or Y; so he just did all of this work and yet that's it.
他不会返回A或B或temp,肯定也不是X或;,所以他刚才做了所有的工作,就是那样。
And you can very quickly take this feature to an extreme 10 and start putting X and Y and A and B and 10 and all your variables up top because it would seem to solve all of your problems and stop all of your thinking, but it's generally not a good thing.
你可能迅速地使这个特征成为一个极端,开始把X和Y和A和B和0,和所有的变量都放在最前,因为好像它可以,解决你的所有问题,并中止你的想法,但通常这不是一件好事情。
I'm doing a test there to say, x if the string x is less than the value of b, and x does not appear before b as strings, then I was going to do, oh, a couple of things, because they're at the same block level.
我正在做一个测试来看看,字符串,是不是小于b的值,而且x不在b之前作为字符串出现,如果是真的话我就要做,一系列的操作,因为他们是在同一个块等级上。
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