Chip, a single individual has two sales of stock during the current year. The first sale produces a short-term loss of $10,000 and the second sale results in a long-term gain of $40,000. Chip's taxable income without considering the gain is $150,000. Chip's stock transactions will increase his income tax liability by:()。
A、$3,000
B、$4,500
C、$6,600
D、$7,200
E、$9,000
第1题:
A.<;stock:quote&ensp;symbol=";SUNW";&ensp;/>;${var}
B.${var}.&ensp;<;stock:quote&ensp;symbol=";SUNW";&ensp;/>;
C.<;stock:quote&ensp;symbol=";SUNW";>;.&ensp;${var}.&ensp;<;/stock:quote>;
D.<;stock:quote&ensp;symbol=";SUNW";&ensp;var=";quote";&ensp;/>;${quote}
第2题:
阅读下列函数说明和C代码,将应填入(n)处的字句写在对应栏内。
【说明】
设有一个带表头结点的双向循环链表L,每个结点有4个数据成员:指向前驱结点的指针prior、指向后继结点的指针next、存放数据的成员data和访问频度freq。所有结点的freq初始时都为0。每当在链表上进行一次L.Locate(x)操作时,令元素值x的结点的访问频度 freq加1,并将该结点前移,链接到现它的访问频度相等的结点后面,使得链表中所有结点保持按访问频度递减的顺序排列,以使频繁访问的结点总是靠近表头。
【函数】
void Locate( int &x)
{ <结点类型说明>
* p =first -> next;
while(p!=frist&&(1))P=P->next;
if(p! =first) /*链表中存在x*/
{(2);
<结点类型说明>
* current = P; /*从链表中摘下这个结点*/
Current -> prior -> next = current -> next;
Current -> next -> prior = current -> prior;
P = current -> prior; /*寻找重新插入的位置*/
While(p! =first &&(3))p=p->prior;
Current-> next =(4); /*插入在P之后*?
Current -> prior = P;
P -> next -> prior = current;
P->next=(5);
}
else printf("Sorry. Not find! \n"); /*没找到*/
}
第3题:
●试题三
阅读下列函数说明和C代码,将应填入(n)处的字句写在答题纸的对应栏内。
【说明】
设有一个带表头结点的双向循环链表L,每个结点有4个数据成员:指向前驱结点的指针prior、指向后继结点的指针next、存放数据的成员data和访问频度freq。所有结点的freq初始时都为0。每当在链表上进行一次L.Locate(x)操作时,令元素值x的结点的访问频度freq加1,并将该结点前移,链接到现它的访问频度相等的结点后面,使得链表中所有结点保持按访问频度递减的顺序排列,以使频繁访问的结点总是靠近表头。
【函数】
void Locate(int &x)
{<结点类型说明>
*p=first->next;
while(p!=frist && (1) )P=P->next;
if (p!=first)/*链表中存在x*/
{ (2) ;
<结点类型说明>
*current=p;/*从链表中摘下这个结点*/
Current->prior->next=current->next;
Current->next->prior=current->prior;
P=current->prior;/*寻找重新插入的位置*/
While(p!=first && (3) )p=p->prior;
Current->next= (4) ;/*插入在P之后*?
Current->prior=P;
P->next->prior=current;
P->next= (5) ;
}
else printf("Sorry.Not find!\n");/*没找到*/
}
●试题三
【答案】(1)p->data!=x(2)p->freq++(3)current->freq>p->freq(4)p->next
(5)current
【解析】(1)空所在的循环是定位x,将指针指向x结点(如存在的话),因此(1)空应填写"p->data!=x"。显然,(2)空是使该结点的访问频度加1,因此(2)空应填写"p->freq++"。(3)空所在的循环是根据访问频度定位x结点的新位置,用P指向x结点的前驱,因此(3)空处应填"current->freq>p->freq"。
(4)、(5)空之间的语句是将结点x插入在P之后。(4)空所在语句是将指针P指向x结点的前驱,因此(4)空应填写"p->next"。(5)空所在语句是将P后继指向结点current,因此空(5)处应填写"current"。
第4题:
A.GP(32chip)+SYNC(64chip)
B.GP(32chip)+SYNC(128chip)
C.SYNC(64chip)+GP(32chip)
D.SYNC(128chip)+GP(32chip)
第5题:
A. The ONS 15216 OADM can be provisioned to add/drop one, two, or four channels via Cisco Transport Manager (CTM).
B. The ONS 15216 OADM consists only of a four-channel module, and the number of ports utilized determines whether it is a one, two, or four channel OADM.
C. Three separate OADM modules are available for one, two, and four channels. The wavelengths to be dropped and added are programmable via Cisco Transport Manager (CTM).
D. The ONS 15216 OADM consists only of a single-channel module. Two-channel and four-channel support is available by stacking the single-channel module.
E. The ONS 15216 OADM is available in one, two, four channels. There are 32 single-channel OADM modules, 16 two-channel OADM modules, and 8 four-channel OADM modules.
第6题:
A.Has-arelationshipsshouldneverbeencapsulated.
B.Has-arelationshipsshouldbeimplementedusinginheritance.
C.Has-arelationshipscanbeimplementedusinginstancevariables.
D.Is-arelationshipscanbeimplementedusingtheextendskeyword.
E.Is-arelationshipscanbeimplementedusingtheimplementskeyword.
F.Anarrayoracollectioncanbeusedtoimplementaone-to-manyhas-arelationship.
G.TherelationshipbetweenMovieandActressisanexampleofanis-arelationship.
第7题:
A.TheONS15216OADMcanbeprovisionedtoadd/dropone,two,orfourchannelsviaCiscoTransportManager(CTM).
B.TheONS15216OADMconsistsonlyofafour-channelmodule,andthenumberofportsutilizeddetermineswhetheritisaone,two,orfourchannelOADM.
C.ThreeseparateOADMmodulesareavailableforone,two,andfourchannels.ThewavelengthstobedroppedandaddedareprogrammableviaCiscoTransportManager(CTM).
D.TheONS15216OADMconsistsonlyofasingle-channelmodule.Two-channelandfour-channelsupportisavailablebystackingthesingle-channelmodule.
E.TheONS15216OADMisavailableinone,two,fourchannels.Thereare32single-channelOADMmodules,16two-channelOADMmodules,and8four-channelOADMmodules.
第8题:
A、is, to translate
B、has, to translate
C、is, in translating
D、has, in translating
第9题:
A.GP(32chip)+SYNC(64chip)
B.GP(32chip)+SYNC(128chip)
C.SYNC(64chip)+GP(32chip)
D.SYNC(128chip)+GP(32chip)
第10题: