• In each case,the Republicans won, by a vote of eight to seven.

    VOA: special.2010.03.04

  • OK, in each case, what these things are doing, is they're doing, what sometimes gets referred to as operator overloading.

    这将会是个默认的或者说是,通用的来比较对象是否相同的方法,好,在每个例子中,这些事情正在做的,就是,一些与操作符重载有关的东西。

    麻省理工公开课 - 计算机科学及编程导论课程节选

  • So we know that in each case the heat is going to be the opposite of the work, but the work isn't the same in these two different ways of getting from here to here, right. So let's just see it explicitly. Here's our qA.

    所以我们知道在每种情形下功,与热量相差一个负号,但从这里到这里,在这两条路径,中的功是不同的,对吧?,那么让我们明确地看一下。

    麻省理工公开课 - 热力学与动力学课程节选

  • And then confronted with a new case, we found ourselves reexamining those principles, revising each in the light of the other.

    当我们面临新的情况时,我们重新检验这些原则,根据新的情况修正这些理由或原则。

    耶鲁公开课 - 公正课程节选

  • Two pieces of memory actually touching each other or you touching memory that you don't actually own, in which case the computer doesn't really know what to do and just, bam.

    两块内存相互覆盖,或者你覆盖了不存在的内存,那样的话计算机不知道,该怎么做,只有崩溃。

    哈佛公开课 - 计算机科学课程节选

  • In fact, it took the same number of steps as it did in the other case, because each time I'm cutting it down by a half.

    因为每次我都把问题的规模,缩小一半,这很棒,好,接下来让我们这么做。

    麻省理工公开课 - 计算机科学及编程导论课程节选

  • In the second case, if both firms undercut each other, you end up with low prices, that's actually good for consumers but bad for firms.

    在第二个案例中,如果两家企业互相削弱,最终会压低价格,这对消费者有利,但对企业不利

    耶鲁公开课 - 博弈论课程节选

  • The first grade corresponds to the row player, me in this case, and the second grade in each box corresponds to the column player, my pair in this case.

    单元格内第一个成绩是我的成绩,每个单元格第二个成绩是,我对手的成绩

    耶鲁公开课 - 博弈论课程节选

  • How many moles of gas are there in each case, in reactants and products? If that changes, of course you know that the pressure in there is going to change at constant volume if the amount of gas in there is changing.

    在反应物和生成物中,各有多少摩尔的气体?,如果它发生了变化,当然在等体条件下,如果气体的总量,发生了变化,压强也会发生变化。

    麻省理工公开课 - 热力学与动力学课程节选

  • And in this case, we go from 8 to 4 to 2 to 1 three times and then on each iteration of this algorithm, each pass across the board I'm touching N numbers, so that means I'm doing N things, log N times.

    在这个例子中,我们从8得到4,到2,再到1,是3次,在这个算法的每次迭代中,每一趟我都会操作N个数,也就是所我每次要做N步操作,一共要做,log,N,次。

    哈佛公开课 - 计算机科学课程节选

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