问题 完形填空

根据对话情景的内容,从对话后所给的选项中选出能够填入每一空白处的最佳选项,并将答案填涂在答题卡上。选项中有两个为多余选项。

---Good evening. Grandma’s Restaurant. May I help you?

--- Good evening.   小题1:  

---Thank you. May I have your name, Miss?

---Mary Brown.

---   小题2: 

---I’d like to order the “Free Matching Dinner for 2”.

---Yes, you may choose two soups, two drinks and four main courses from the menu.

---   小题3:  

---Are you ready? What Kind of soup do you want?

---Chicken soup and tomato soup.

---     小题4:  

---I’d like a plate of noodles, two chicken wings, some beef and a bowl of rice.

---    小题5:  

--- Just a coke and some green tea, please.

---May I have your address, please?

---Sure, 6A, Kingston Court, Belair Cardens, Shatin.

---Okay, your food will be delivered in half an hour. Thanks for calling.

A.May I book a table in your restaurant for tonight?

B.I would like to order some food for dinner.

C.Miss Brown, please tell me your order.

D.How about your main courses?E. Would you like some drinks?

F. Okay. Let me see.

G. And your drinks?

答案

小题1:B

小题2:C

小题3:F

小题4:D

小题5:G

题目分析:文章是外卖电话订餐的对话,内容较为简单。

小题1:根据上文,Good evening. Grandma’s Restaurant. May I help you?这里应该填:我想要订餐。所以选B

小题2:根据下文,I’d like to order the “Free Matching Dinner for 2”.此处是询问订餐内容的问话,所以选C

小题3:根据下文,Are you ready? What Kind of soup do you want?这里顾客是要求思考一会儿,所以选F

小题4:根据下文,-I’d like a plate of noodles, two chicken wings, some beef and a bowl of rice.这里填的是询问主食的问题,所以选D

小题5:根据下文,Just a coke and some green tea, please.此处的问题是询问饮料,所以选G

问答题
问答题

请编制程序PROG1.ASM,其功能是:将内存中的20个有符号字节数据依次除以5,并按照四舍五入原则(即余数绝对值的2倍小于除数,则舍去)将商存入指定的内存区域。
例如:
内存中有:10H,01H,27H,00H,FFH,F8H,……
结果为:03H,00H,08H,00H,00H,FEH,……
部分程序已给出,其中原始数据由过程LOAD从文件INPUT1.DAT中读入SOURCE开始的内存单元中,运算结果要求从RESULT开始存放,由过程SAVE保存到文件 OUTPUT1.DAT中。
填空BEGIN和END之间已给出的源程序使其完整(空白已用横线标出,每行空白一般只需一条指令,但采用功能相当的多条指令亦可),或删除BEGIN和END之间原有的代码并自行编程来完成要求的功能。
对程序必须进行汇编,并与IO.OBJ链接产生PROG1.EXE执行文件,最终运行程序产生结果(无结果或结果不正确者均不得分)。调试中若发现整个程序中存在错误之处,请加以修改。
试题程序:
EXTRN LOAD:FAR,SAVE:FAR
N EQU20
X EQU5
DSEG SEGMENT
SOURCE DBN DUP()
RESULT DBN DUP(0)
NAME0 DB’INPUT1.DAT’,0
NAME1 DB’OUTPUT1.DAT’,0
DSEG ENDS
SSEG SEGMENT STACK
DB 256 DUP ()
SSEG ENDS
CSEG SEGMENT
ASSUME CS:CSEC, SS:SSEC, DS:DSEG
START PROCFAR
PUSHDS
XOR AX,AX
PUSHAX
MOV AX,DSEG
MOV DS,AX
MOV ES,AX
LEA DX,SOURCE
LEA SI,NAME0
MOV CX,N
CALLLOAD
; *** BEGIN ***
LEA SI,SOURCE
LEA DI,RESULT
CONT: LODSB
CBW
MOV DL,X
IDIVDL
(1) ; 余数为负数吗
JNS PLUS
(2) ; 求绝对值
ADD AH,AH
CMP AH,DL
JB (3)
SUB AL,1
JMP NEXT
PLUS: ADD (4)
CMP AH,DL
JB NEXT
(5)
NEXT: STOSB
CMP SI,N
JB CONT
; *** END ***
LEA DX,RESULT
LEA SI,NAME 1
MOV CX,N
CALLSAVE
RET
START ENDP
CSEG ENDS
ENDSTART