一般计算机声卡的Line Out接口通过音频线来连接音箱的（ ）接口。

A:Line in B:Mic C:Line out D:Line

We trust the mentioned shipment will reach you（）sound condition.

Passage Three
Did you know that a turtle can lay twelve eggs in one minute A large sea turtle lays around 150 eggs at a time. She lays all these eggs in just a few minutes.
Large sea turtles live in the warm seas of the world. Except for when they lay their eggs, they spend their whole lives in water. When it is time to lay their eggs, the females swim to land. They usually return to the place where they themselves were born. How they find their way back there is a mystery.
When they reach shore, the big, heavy turtles crawl slowly up to the high water mark. Using their flippers, they pull themselves along the sand. They struggle like mountain climbers to attain their goal. When they finally reach dry sand, they rest before beginning the difficult task of laying eggs.
The turtles lay the eggs in deep holes and cover them with warm sand. The sand protects the eggs from harm. Then the females leave them. After a few weeks, if you happened to be walking along the beach, you might see the sand begin to shake in one spot. Then you would see tidy black balls coming out of the sand. The tidy heads of baby turtles!
Baby turtles have a build-in sense of direction. As soon as they are hatched, they head for the water. Once the babies swim out to sea, they don’t touch shore again until it is time for them to lay their own eggs.

A:see B:push C:bury D:reach

fclose (out);

void isSort()
{
int i, j, data;
for(i=0;i＜AII;i++) ／*对数组中的D位数按后C位进行从大到小的排序*／
for(j=i+A; j＜B00; j++)
{
if (aa[i)%A000＜aa[j]%A000)/*比较每个数的后C位*/
{data=aa[i]; aa[i]=aa[j]; aa[i]二data；)
else if(aa[i]%A000==aa[j]%A000) ／*若后C位数相等*／
if (aa[i]＞aa[j] ／*则要按原D位数的值进行从小到大的排序*／
{data=aa[i]; aa[i]=aa[j]; aa[j]=data; )
}
for(i=0; i＜A0; i++) ／*将排序后的前A0个数存入数组b中*／
bb[i]=aa[i];
}
[解析] 本题实质上考的是对数组元素进行排序的问题。最容易理解的方法是选择法。选择法的基本思想是：依次用当前取得的元素和它后面的其他元素进行比较，比如要从大到小进行排序，则只要发现后面的元素中有比当前元素大的，就把刚发现的较大的元素与当前元素交换，这样做的目的是当与其后的所有元素都比较完后，可以保证当前位置的当前元素存放着最大数。按照同样的方法去取得次大数。由于本题中题目要求要对数组中的4位数的后3位进行排序，因此，要首先取得各个数的后3位并做比较，“aa[i]%1000”就可以得到每个数的后3位。根据题意，对这些数按后3位进行从大到小的排序。若后3位数相等，则要按原4位数的值进行从小到大的排序。

fclose(out);

void jsSort()
{
int i, j,data;
for(i=0;i＜AII;i++)
for(j=i+A;j＜B00;j++)
{
if(aa[i]%A000＞aa[j]%A000) /*按照每个数的后C位的大小进行升序排列*/
{
data=aa[i];
aa[i]=aa[j];
aa[j]=data;
}
else if(aa[i]%A000==aa[j]%A000 /*如果后C位数值相等，则对这些数值按原始D位数据
进行降序排列* /
if(aa[i]＜aa[j])
{
data=aa[i];
aa[i]=aa[j];
aa[j]=data;
}
}
for(i=0;i＜A0;i++) /*将排好序的前十个数存入数组bb中*/
bb[i]=aa[i];
}[解析] 本题考查的知识点如下：
(1) 循环结构的嵌套。
(2) 选择结构的嵌套。
(3) 特殊运算符“%”的使用。
(4)数据的升降序排列。
在该题中，首先要解决的问题是如何把题中4位数转换成符合要求的3位数。这里要用到一个比较特殊的运算符%，它的功能是取余。将4位数除以1000取余则得到要求的3位数。根据题意，若后3位相等，则按照原数据进行降序排序，所以在这里要加一个选择结构进行判断。对数据进行排序，可以使用最简单的“选择排序法”。即第一次从第一个数据开始和所有后面的数据进行比较，与比它大(小)的数据交换位置，将数据中最大(小)的数交换到最前面。第二次从第二个数据开始与后面的数据进行比较，将此次比较中最(小)的数据交换至第二个位置，直至最后一个数据。

fclose(out);

void jsSort()
{ int i，j，k，strl；
char ch；
for(i=0；i＜20；i++)
{ strl=strlen(xx[i])； /*求各行字符串的长度*/
for(j=1;j＜strl-2;j=j+2) /*将下标为奇数的字符按其ASCII值从小到大的顺
序进行排序*/
for(k=j+2；k＜strl；k=k+2)
if(xx[i][j]＞xx[i] [k])
{
ch=xx[i][j]；
xx[i][j]=xx[i] [k]；
xx[i][k]=ch；
}
}
}