C1 be the right-most character of s1.
C1是S1最右侧的字符。
The first part, S1..., that's easy. Sn.
第一部分是,S1,这很容易,Sn
Record a script called S1, and build it.
记录名为s1的脚本，并构建它。
S 1：多少男孩？
So, the address is 71 so what ends up in s1?
地址是71，那么最终s1是什么？
For example, schema_map ((s1, t1), (s2, t2)).
例如，schema_map ((s 1,t1)，(s 2,t2))。
4gl removes it and replaces it with Spaces denoted by s1.
4gl删除它，并用s1表示的空格代替它。
The base case is whenever S1 or S2 is a zero-length string.
（基线条件（basecase）是S1或S2为长度为零的字符串的情形。
This statement is represented in the example as subquery S1.
这个语句在示例中表示为子查询s1。
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。
If it's okay, convert the remaining chars in s1 into an integer.
如果没有问题，则将s1中的余下字符转换成整数。
In this case, the LCS of S1 and S2 is clearly a zero-length string.
在这种情况下，S1和S2的LCS显然是长度为零的字符串。
Then, we would have to add a NULL character after the last char copied into s1.
然后，我们可以在复制到s1的最后一个字符后添加NULL字符。
The second aggregation function is applied to the partially calculated results of S1.
第二个聚合函数应用于S1的部分计算结果。
Now the functional test script, S1, is associated with the TestManager test case, TC1.
现在功能性测试脚本，S1，就与TestManager测试用例，TC1联系在一起。
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。
Well, let's think about comparing the flux integral for S1 and the flux integral for S2.
让我们考虑关于S1和S2的通量积分的比较。
Then we check that the first two chars of s1 are "-s", to indicate the start page option.
然后我们检查指示起始页选项的s1的头两个字符是否为“-s”。
Since you suspended the change to S2, you have version S1 selected again in your workspace.
由于您悬挂了到S2的变更，所以您的工作区中再次选择了版本s1。
So, if we now apply the logic from today, char * s1 means s1 A pointer or the address of what?
所以，如果我运用这个逻辑，char，*s1的意思是,不是一个字符型，而是什么？，that，s1，is，not，a，char，，it’s，instead，what？,一个指针或者什么的地址？
Next, you need to obtain the actual alignment strings - S1 'and S2' - and the alignment score.
接下来，需要得到实际的比对字符串—s1'和S2 '—以及比对的得分。
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。
In the example, this is represented by the outermost select statement, which USES S1 as a subquery.
在这个示例中，这由最外层的选择语句表示，这个语句将S1用作子查询。
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。
So, in Listing 6, step S0 is executed followed by steps S1 through S4, which are executed simultaneously.
所以，在清单6中，执行完步骤s0之后执行步骤s1到S4，它们是同步执行的。
Select the functional test script name, S1, which you need to associate with the TC1 test case, and click OK.
选择功能性测试脚本名，S1，您需要将其与TC1测试用例相联系，并点击OK。
s1 Here's s1, it's pointing to this array, bracket I is bracket zero initially so it's pointing at capital F. Same deal.
i】是什么？，Well，，what’s，s1，bracket，I？,这里是s1，指向的是这个数组，【i】开始是【0】，它指向的是,大写的F，同样的道理。
s1 Here's s1, it's pointing to this array, bracket I is bracket zero initially so it's pointing at capital F. Same deal.
i】是什么？，Well，，what’s，s1，bracket，I？,这里是s1，指向的是这个数组，【i】开始是【0】，它指向的是,大写的F，同样的道理。
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