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Suppose f of x, y, z equals k1, that is my equation, s1 and it gives me a solution s1.
假设我的方程是这样，然后给出了一个解。
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So in the end, this is just saying allocate me as many bytes s1 as were needed to store s1 itself.
最终，这个只是表示给我分配,足够多的空间来存储。
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s1 Here is 1s atomic. But lithium has 2s1, so I need a 2s atomic orbital here and likewise over here.
这是1s原子，但锂有两层，所以我还需要在这里添加2s轨道,就像那样。
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And none of this has an S1 in it, so that all goes away.
1在这里面就都消掉了
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s1 So for the fourth period, now we're into the 4 s 1 3d for potassium here. And what we notice when we get to the third element in 4s2 and the fourth period is 3d that we go 4 s 2 and then we're back to the 3 d's.
对于第四周期到现在我们来到钾的1,然后我们返回到,我们注意到当我们看到第三个元素,第四周期我们来到,然后我们返回到。
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But I can also ask for parts of this. So I can, for example, say give me the first element of s1 string 1, s 1. Ah, that's exactly what we would have thought if this was represented as an ordered sequence of things.
然后返回一个长字符串，我也可以，取字符串的一部分，例如,我可以要求返回给我字符串,的第一个元素，啊，这证明我们之前,认为的它是一个有序的字符序列的想法。
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And maybe S1 through Sn results in the destruction of my soul.
也许是S1到Sn导致了灵魂的毁灭
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s1 I have a variable called s1 and it's char * of type char * so here we go.
我有一个变量,它的类型是。
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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】
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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。
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char * s1 Then I go ahead and declare a char * called s1.
然后我声明。
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What we're going to do is, we're asking the question, what is the maximum, choosing S1, of this profit.
下面我们要问问我们自己,选S1下利润的最大值是多少呢
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char *s2 = s1 >> David: Correct, yep, char *s2 gets s1.
>，大卫：正确的，是。
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s1 We're assigning to s1 the string that the user typed in.
我们把用户输入的字符串赋值给。
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char *s2 = s1 It's char * s2 gets s1.
就是。
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To make this a first order condition, I'm going to say "at the best response," put a hat over the 1.
为了达到一阶条件,我说在最佳对策下,在S1上写个帽
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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
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This is copy2.c. At the very beginning I, again, demand say something and then I declare s1 to be a string, aka char *, and I store in s1 the string the user types in.
这是copy2。c，在开头打印一句话，然后声明s1是一个字符串，也叫做char，*，然后把用户输入的字符串存储在s1中。
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So the profit is 2 S1 plus S2 plus B S1 S2 minus S1 squared.
利润是2-S1平方
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Again if I go back here, let me comment this out.
一对字符串，s1和s3，让我们运行下他们。
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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。
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Let's just assume that s1 is a pointer and as an arrow suggests it's pointing at this byte here.
我们假设s1是一个指针,就像箭头所表示的，指向这个字节。
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I store in s1 the address of the first byte that the user typed in and by the way that first bite happens to live in this new place called the heap and that's the only update to the story thus far.
我在s1中存储的是用户输入的字符串的首地址,这样第一个字节存储在这块,新的堆的地方,这个是唯一的修正。
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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？
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So, after the second line for code here char * s1 gets the return value of get string this is what the state of our world looks like.
在第二行之后,这个代码char，*s1，等于GetSting的返回值,这就是它看起来的状态。
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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
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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？
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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
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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？
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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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