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 ver,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(僅供參考)
1 翻譯題(必答)
CDMA 廠商們已經竭力開發出一種用於CDMA手機上使用的名爲RUIM的SIM卡,進而能夠使CDMA漫遊,這種爲中國聯通開發的新CDMA上使用的SIM卡在 中國進一步開發與推廣,韓國KTF公司在今年年初向公衆展示其採用此卡在GSM與CDMA間進行漫遊,儘管如此, 也只有那些含有用戶服務數據的卡才能實現漫遊功能, 相反那些一般的CDMA手機或只含有用戶手機號的是不行的(除非通過轉移)
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. 編程題(必答)
鏈表
a 實現一個整數鏈表, 並支持插入運算(在給定的結果後插入)和刪除運算(移除給定的結點),
b 實現一個方法來對鏈表進行除序排序;
3. Debugging (Mandatory)
a. For each of the following recursive (遞歸)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
