下面是广东北늚W试题(中英文题Q,q套题早已在|络上流传数q_从来只见题目Q不见解{,

一Q英文题?br />1. Tranlation (Mandatory)
CDMA venders have worked hard to give CDMA roaming capabilities via the development of RUIM-essentially, a SIM card for CDMA handsets currently being deployed in China for new CDMA operator China Unicom. Korean cellco KTF demonstrated earlier this year the ability to roam between GSM and CDMA using such cards.However,only the card containing the user’s service data can roam-not the CDMA handset or the user’s number (except via call forwarding).
译QCDMA开发商一直致力于RUIM卡的开发,以此赋予CDMA漫游的能力。RUIM卡类gSIM卡,事实上目前它已经被中国的CDMAq营商中国联通广泛用。韩国手机制造企业KTF今年早些时候展CZ使用此种卡在GSM和CDMA|络中O游的功能Q但是,只有该卡包含的用h务数据能够O游,CDMA手机本n及用户号码则不能Q除了呼叫前转业务)?br />呵呵。上文可能翻译的不太_ևQ欢q批评?/font>
2. Programming (Mandatory)
Linked list
a. Implement a linked list for integers,which supports the insertafter (insert a node after a specified node) and removeafter (remove the node after a specified node) methods;
b. Implement a method to sort the linked list to descending order.
{:题目的意思是实现一个整型链表,支持插入Q删除操作(有特D要求,都是在指定节点后q行操作Q,q写一个对链表数据q行降序排序的方法?br />那我们不妨以一个线性链表进行编E?br />
// 单链表结构体?br />typedef struct LNode
{
int data;
struct LNode *next;
}LNode, *pLinkList;
// 单链表类
class LinkList
{
private:
pLinkList m_pList;
int m_listLength;
public:
LinkList();
~LinkList();
bool InsertAfter(int afternode, int data);//插入
bool RemoveAfter(int removenode);//删除
void sort();//排序
};
实现Ҏ
//insert a node after a specified node
bool LinkList::InsertAfter(int afternode, int data)
{
LNode *pTemp = m_pList;
int curPos = -1;
if (afternode > m_listLength ) // 插入点超q总长?br /> {
return false;
}
while (pTemp != NULL) // 扑ֈ指定的节?br /> {
curPos++;
if (curPos == afternode)
break;
pTemp = pTemp->next;
}
if (curPos != afternode) // 节点未寻刎ͼ错误退?br /> {
return false;
}
LNode *newNode = new LNode; // 新节点插入指定节点?br /> newNode->data = data;
newNode->next = pTemp->next;
pTemp->next = newNode;
m_listLength++;
return true;
}
//remove the node after a specified node
bool LinkList::RemoveAfter(int removenode)
{
LNode *pTemp = m_pList;
int curPos=-1;
if (removenode > m_listLength) // 删除点超q总长?br /> {
return false;
}
// 扑ֈ指定的节点后一个节点,因ؓ删除的是后一个节?br /> while (pTemp != NULL)
{
curPos++;
if (curPos == removenode+1)
break;
pTemp = pTemp->next;
}
if (curPos != removenode) // 节点未寻刎ͼ错误退?br /> {
return false;
}
LNode *pDel = NULL; // 删除节点
pDel = pTemp->next;
pTemp->next = pDel->next;
delete pDel;
m_listLength--;
return true;
}
//sort the linked list to descending order.
void LinkList::sort()
{
if (m_listLength<=1)
{
return;
}
LNode *pTemp = m_pList;
int temp;
// 选择法排?br /> for(int i=0;i<m_listLength-1;i++)
for(int j=i+1;j<m_listLength;j++)
if (pTemp[i].data<pTemp[j].data)
{
temp=pTemp[i].data;
pTemp[i].data=pTemp[j].data;
pTemp[j].data=temp;
}
}
前两个函数实C要求aQ后一个函数sort()实现了要求b
3. Debugging (Mandatory)
a. For each of the following recursive methods, enter Y in the answer box if the method terminaters (assume i=5), Otherwise enter N.
Q题目意思:判断下面的递归函数是否可以l束Q?br />
static int f(int i){
return f(i-1)*f(i-1);
}
static int f(int i){
if(i==0){return 1;}
else {return f(i-1)*f(i-1);}
}
Ansewr:YQ当iQ?时可l束递归
static int f(int i){
if(i==0){return 1;}
else {return f(i-1)*f(i-2);}
}
b. There are two errors in the following JAVA program:
static void g(int i){
if(i==1){return;}
if(i%2==0){g(i/2);return;}
else {g(3*i);return;}
}
3 10 5 16 8 4 2 1
{:在第一个if语句前加 System.out.print(i+" ");
else 里面的g(3*i)改ؓg(3*i+1)
该题q友alvin补上Q我不熟java。谢谢他。
Q-Q-Q-Q-Q-Q-Q-Q-Q-Q-Q-Q-Q-Q-Q-Q-Q-Q-Q-Q-Q-Q-Q-
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中文W试?/font>
1Q汉译英
北电|络的开发者计划来自于不同组l的开发者,能够在北늽l的q_上开发圆满的补充业务。北늽l符合工业标准的开放接口,充业务的开展引入了无数商机Q开发者计划ؓ不同层面的开发者提供不同等U的资格Q资格的划分q考虑C下因素:补充业务与北늽l^台的集合E度Q开发者团体与北电|络的合作关p,{等?br />{:呵呵。这个这个基本上q是不现丑了吧?img title="cry" style="BORDER-TOP-WIDTH: 0px; PADDING-RIGHT: 2px; PADDING-LEFT: 2px; BORDER-LEFT-WIDTH: 0px; BORDER-BOTTOM-WIDTH: 0px; PADDING-BOTTOM: 2px; PADDING-TOP: 2px; BORDER-RIGHT-WIDTH: 0px" height="48" alt="cry" src="http://goodsoft.cn/blog/image/face/cry.gif" width="48" />
2Q编E?/font>
整数{换成字符Ԍvoid itoa(int,char);
例如itoa(-123,s[])则s=?123?
{:
char* itoa(int value, char* string)
{
char tmp[33];
char* tp = tmp;
int i;
unsigned v;
char* sp;
// D{为正?br /> if (value < 0)
v = -value;
else
v = (unsigned)value;
// 数转换为字W放在数ltmp?br /> while (v)
{
i = v % 10;
v = v / 10;
*tp++ = i+'0';
}
// tmp里的字符填入string指针里,q加上负P如果有)
sp = string;
if (value < 0)
*sp++ = '-';
while (tp > tmp)
*sp++ = *--tp;
*sp = 0;
return string;
}