The first step is divide them in half. All right? I'm not sure how to mark it here, remember I need to come back there. I'm not yet done. What do I do?
我不能确定如何,在这里进行标记,记住我会回这里进行讲解,我还没做完呢?
But I want to stress again, as long as I do the base case right and my inductive or recursive step reduces it to a smaller version of the same problem, the code will in fact converge and give me out an answer.
就开心的去做吧,但是我想再次强调,只要基础事件处理正确而我的递归,或递推步骤能把它简化为更简单的同类问题,那么这段代码就可以收敛。
Now q2 was in this step, and we're going to leave that reversible, right, but q1 is irreversible.
现在q2是这个过程中的热,是一个可逆的,但是q1是不可逆的。
On the next step though, this, we get substituted by that. Right, on the next step, I'm back in the even case, it's going to take six more steps, plus t of b minus 1. Oops, sorry about that, over 2.
这一步就是偶数了,这一步会让我们得到,6+t这样的结果,因为b-1现在是偶数了,别忽略这里的细节。
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