So at this point in the story, we're not gonna forge ahead to here or to here, or the last two lines.
那么这时,我们并不能继续执行,这行或这行,或者说后两行的指令。
If you look really carefully, you could find it was not a single line but two lines, and Bohr is silent on such matters.
如果你仔细观察的话,会发现这是两条谱线而非一条,对这一研究,波尔不得不沉默。
There are no less than twenty pages devoted to the critical debate on these two lines: what the hell Saint Peter's two-handed engine is, what it looks like, what it does, what -- and everything about it.
集中描写这两行里的争论的,不少于20页:,圣彼得的双手引擎到底是什么,长什么样,用来干什么-和关于它的所有所有。
That's just the those-- let me use my pointer that's just these two lines here. I checked the value, and in one case, I'm changing the last to be mid minus 1, which is the case I'm in here and I just call again. All right?
来看看ppt,这里有两条线,我检查了值,在一种情况下,我把last指向了中点减一,就是这种情况,我们再来调用一下怎么样?
So what we're going to do is take the equations for those two lines, so here's one of those equations and here's the other one, set the P in those equations equal to X, I've got two equations in one unknown, I'm sorry, I've got one equation and one unknown.
接下来我们只需要,列出这两条线各自的方程,也就是这个方程和这个方程,把方程中的P换成X,我就得到了两个等式和一个未知数,错了,是一个等式和一个未知数
So input is an int called F, print out to the screen or your piece of paper the value of C. Odds are you only need two, maybe three lines of code for this, but the goal ultimately will be try it on your own.
所以输入是一个int类型的F,在你的屏幕上,或你们的纸上打印出C的数值,可能你们需要两行或者三行的代码,但是最终的目标需要你自己独自尝试。
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