This indicates that by combining two lithiums, the energy of the combined system is lower than the energies of the atomic systems.
结果显示,两个锂结合后,其结合后的能量,比原子体系的能量要低。
And those instructions, by the way, are very simple: they're things like, take the value out of two places in memory, and run them through the multiplier in here, a little piece of circuitry, and stick them back into someplace in memory.
顺便说一句,那些指令非常简单:,他们从记忆存储器中,取出两个位置的结果,在倍增器中运行,一小部分电路,再将它们插回存储器中的某些地方去。
If I stop, if I move slowly, if I move more slowly then these two will want equilibrium.
如果我停下来,如果我缓慢地移动活塞,非常缓慢,结果这两个压强会平衡。
How many bonding electrons does c l have? All right. Let's see, we've got a mixed response here, it turns out it has two bonding electrons.
氯有多少成键电子?好,让我们来看看,大家的回答不太一致,结果应该是两个成键电子。
Well, it turns out that if you want to check two conditions and you only care that one of them is true or the other one is true.
好的,结果是如果你想要核对两个条件,你只关心其中一个是正确的,或另外一个是正确的。
Here is the explanation of why those 2 things work differently in the reassembling cases.
现在我给你们讲解一下为何,这两个重组例子的结果会不同。
So, suppose I tell you to add two vectors, A and B equal to C, and I say, "What's the result of adding A and B?"
假设我告诉你们将两个矢量相加,A + B = C,我问,"A 加 B 结果是什么"
ve matched to that, and two observations.
我们做到了,另外有两个观察结果,So,,we’
Do one comparison and return one of two possible orders on it, but I need to decide that.
虽然可以做一次比较,然后返回两个可能的顺序中的一个为结果,但是这需要我来做决定。
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。
So we'll start to look at molecules and we'll see if we take two atoms and we fill in our molecular orbital and it turns out that they have more anti-bonding orbitals than bonding, that's -- a diatomic molecule we'll never see.
我们要看开始看一看分子,并且我们会发现如果我们,取两个原子并且填入分子轨道,结果是它们的反键轨道,比成键轨道更多,这就是-一个我们不会看到的二元子分子。
It says, well I'm going to print out first and last just so you can see it, and then I say, gee 2 if last minus first is less than 2, that is, if there's no more than two elements left in the list, then I can just check those two elements and return the answer.
然后它计算了尾点和开始点的差,如果小于2的话,也就是说数组中的元素小于等于,我对这两个元素进行比较,然后返回结果就可以了,否则的话,我们就去寻找中值点,注意它是怎么实现的,首先这个指向一个列表的开头。
And if you just look at the numbers, you can see how it cuts in from one side and then the other side as it keeps narrowing that range until it gets down to the place where there are at most two things left, and then it just has to check those two to say whether it's there or not.
你能看到他是如何不断的,从一个大的范围被拦腰劈开,知道最终只剩下两个数字,然后就只需要,比一比就知道结果了,将它同线性查找比较下。
There are two possible values here, I think.
有两个可能的结果值了,我认为。
So I can instead of the dV here, I can insert this R over p dT.
两个p消掉了,结果。
All right? And if you want to do other things, whoops, sorry-- and add them together, we will get out, again, a concatenation of that string. And these will, we'll let you work through the variations, but these are the simple expressions we can use.
对不对?如果你想要进行一些别的操作,哦,对不起-是加到一块,然后结果值也是,两个字符串的链接,这些会,我们会让大家慢慢学习变量,但是这些是我们能用的最简单的表达式。
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