s2 2s2 2p6 3s2 3p5 Chlorine is 1s2, 2s2, 2p6, 3s2, 3p5.
氯的电子结构是。
s2 3p5 That gives us 14, and that's 3s2, 3p5, and so we've got to come up with, we have 32 electrons that we have to use.
则是14,为,因此我们能想到的是,总共需要32个电子。
So this has the effect of starting at location zero and I do s2 bracket I gets s1 bracket I.
这个是从地址0开始,接着,s2【i】,=,s1【i】
So for each possible choice of S2, I'm going to draw Player I's best response and we'll do it in red.
对于任意定义域内的S2,我用红色来表示参与人I的最佳对策
Or, especially when talking about estimates of the variance, we sometimes say S2 or we say standard deviation2.
又或者,在讨论方差估计的时候,我们常用S2,称为标准差的平方
s2 2s2 2p2 We also know that carbon is 1s2, 2s2, 2p2, and we know what hydrogen is.
我们也知道C是,我们也知道H是什么。
s1+s2 Instead I can write with impunity the solution will be s1 plus s2.
而我会写下解就是,毫无风险。
s2 And, in this case, I start with helium -is 1s2.
在这种情况心爱,氦开始的是。
char *s2 = s1 >> David: Correct, yep, char *s2 gets s1.
>,大卫:正确的,是。
f Well, call toupper pass this lowercase F F to this function called toupper it's going to return capital F and so what do I assign to s2 bracket zero?
调用toupper函数,传递这个小写,然后返回的是大写的,那么我对s2【0】赋值多少?
So if Player I chooses S1*, Player II will want to choose S2* since that's her best response.
所以当参与人I选择S1*时,参与人II就会选择最佳对策S2
char *s2 = s1 It's char * s2 gets s1.
就是。
If Player II is playing S2*, Player I will want to play S1* since that's his best response.
如果参与人II选择S2,参与人I就会选最佳对策S1
s2 So, I'm passing in the first character in s2, I'm making it uppercase and then I'm putting it back so casually speaking this is just capitalizing the first letter of whatever word the user typed in to s2.
那么,我把第一个字符传递给,我把它转换成大写的,然后存下来,简单说,这就是把用户输入s2中的单词,的第一个字母转换成大写的。
My algebra, which is often wrong, suggests that the solution is S1* equals S2* equals 1 over .
我代数总是算错,最后解是S1*=S2*=1/
So that is actually making a copy of the address and putting it in s2.
这实际上是复制一个地址,然后放置在s2中。
This term is the same except we're going to subtract off Player iI's efforts squared: S2 squared.
表达式也是一样的,只不过我们要减掉S2的平方
s2 2s1 1s2, 2s1 over here. The same thing, 1s2, 2s1.
把1s2,2s1都填上,同样的。
s2 This thing is called s2 so let's just label it s2 s2 = s1 and now s2 gets s1.
这个东西叫做s2,我们把它标记为2,现在。
It tells me that the best response to S2 is the ?1 that solves this equation, that solves this first order condition.
我们得出S2的最佳对策是?1,?1是这个方程的解,它满足一阶条件
At this point in the story both s1 and s2 are literally pointing at the same location in memory so that's now what the story looks like.
在这里,s1和s2指向的是,内存中同一个地方,这里看起来像什么?
So, in particular, if you're person 1 and then "s-i" would be "s2, s3, s4" up to "sn" but it wouldn't include "s1."
具体来说如果你是参与人1那么,si-可能表示s2,s3,s4,直到sn,但是不包括s1
So, in other words, if we apply some of the storyline 4 5 6 from the previous tale, if s1 happens to be 4, 5, s2 4 5 6 you're just asking the question is 4, 5, 6 equals equals to 1, 2, 3?
换句话说,如果我们运用之前的一些情节,如果s1正好为,正好为1,2,3,6,and,s2,happens,to,be,1,,2,,3,你就会问:,等于1,2,3吗?
So taking advantage of the fact that we know S1 is equal to S2 I can simplify things by making S1* equal to S2*.
我们利用S1=S2这个等式,这样用S1*=S2*可以化简了
And when I solve it for k2 boundary conditions, s2 I get s2.
在k2条件下我,得到解。
When you're working on your homework assignments, if your product, the thing you hand in was just S1 + S2, then you might think what?
比如你们两个人交一份作业,你们交上来的只是S1+S2,那你会怎么想
So the profit is 2 S1 plus S2 plus B S1 S2 minus S1 squared.
利润是2-S1平方
So the very last two lines of code I'm saying this, s2 the original string is s1, the copy of the string is s2, but if I haven't lost you, what are we really going to see when we print this?
最后两行代码是这样的,原字符串是s1,那个字符串的拷贝是,但是如果我没有忘记,当我们打印的时候,我们将看到什么?
If s2 equals s1, guess which character you're also changing?
如果s2,=,s1,猜猜哪个字符将会改变?
Actually, we can do it a little better than that, since we know the game is symmetric, we know that S1* is actually equal to S2*.
实际上我们能得出更多,因为我们知道这个博弈是对称的,我们知道S1*=S2
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