笔试题及答案1

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　　笔试题及答案1

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). 作答(仅供参考)
1 翻译题(必答)
CDMA 厂商们已经竭力开发出一种用于CDMA手机上使用的名为RUIM的SIM卡,进而能够使CDMA漫游,这种为中国联通开发的新CDMA上使用的SIM卡在 中国进一步开发与推广,韩国KTF公司在今年年初向公众展示其采用此卡在GSM与CDMA间进行漫游,尽管如此, 也只有那些含有用户服务数据的卡才能实现漫游功能, 相反那些一般的CDMA手机或只含有用户手机号的是不行的(除非通过转移)
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. 编程题(必答)
链表
a 实现一个整数链表, 并支持插入运算(在给定的结果后插入)和删除运算(移除给定的结点),
b 实现一个方法来对链表进行除序排序;
3. Debugging (Mandatory)
a. For each of the following recursive (递归)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 J***A program:
static void g(int i)
{
if(==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