广东北电的笔试题

时间:2021-01-16 16:30:54 笔试题目 我要投稿

广东北电的笔试题

  1.汉译英

广东北电的笔试题

  北电网络的开发者计划使来自于不同组织的开发者,能够在北电网络的'平台上开发圆满的补充业务。北电网络符合工业标准的开放接口,为补充业务的开展引入了无数商机,开发者计划为不同层面的开发者提供不同等级的资格,资格的划分还考虑到以下因素:补充业务与北电网络平台的集合程度,开发者团体与北电网络的合作关系,等等。

  答:呵呵。这个这个基本上还是不现丑了吧。

  2.编程

  将整数转换成字符串:void itoa(int,char);

  例如itoa(-123,s[])则s=“-123”;

  答:

  char* itoa(int value, char* string)

  {

  char tmp[33];

  char* tp = tmp;

  int i;

  unsigned v;

  char* sp;

  // 将值转为正值

  if (value < 0)

  v = -value;

  else

  v = (unsigned)value;

  // 将数转换为字符放在数组tmp中

  while (v)

  {

  i = v % 10;

  v = v / 10;

  *tp++ = i+'0';

  }

  // 将tmp里的字符填入string指针里,并加上负号(如果有)

  sp = string;

  if (value < 0)

  *sp++ = '-';

  while (tp > tmp)

  *sp++ = *–tp;

  *sp = 0;

  return string;

  }

  英文笔试题

  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).

  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.

  3. Debugging (Mandatory)

  a. For each of the following recursive methods,enter Y in the answer box if themethod terminaters (assume i=5), Otherwise enter N.

  static int f(int i){

  return f(i-1)*f(i-1);

  }

  Ansewr:

  static int f(int i){

  if(i==0){return 1;}

  else {return f(i-1)*f(i-1);}

  }

  Ansewr:

  static int f(int i){

  if(i==0){return 1;}

  else {return f(i-1)*f(i-2);}

  }

  Ansewr:

  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;}

  }

  please correct them to make sure we can get the printed-out result as below:

  3 10 5 16 8 4 2 1

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