1991-09-20 — Page 29

(29)育教:日三十月八年未辛磨夏

報日僑單

五期星:日十二月九(一九九一)年十八國民奉中

日三十月八

4. (á) (1)

Type bit

0.

is an integer.

Line Number

00001112

x 2 + 1 x

230

Correction(s) Required:

changes to 230 PRINT "HOBBIES ARE:"MHOB1$;" MH0B2$;",";Μ/083$

1991中學會考試題建議答案

Computer

270

The bit pattern is 7.

changes to 270 INPUT "CONTINUE(Y/N)";R$

明德出版社,

(ii) Type bit zi

it is a character.

280

-changes to

280 IF R$="N" THEN END

MILL & DALE PRESS

1100000 1x 2 + 1 x

=-96

295

insert, line 295 RESTORE

SUGGESTED ANSWER TO COMPUTER STUDIES, PAPER I

The ASCII code of 96 is. " (iii) Type bit = 0, .. it is an integer.

101111 is a negative number

340

(a)(i)

(iii) Plotter

(iv) Magnetic Tape-

(v)Daisy Wheel Printer"

Bar Code Reader

(ii) Magnetic Ink Reader

430

1011112

X + 1x 2 +

changes to 340 DATA CHAN, JACK",29,M,TENNIS;

SWIMMING, BADMINTON

changes to 430 IF C$="a" AND C$<="2" THÊN"

G$=CHR${ASC(C$)-32).

-32+ 8 + 4 + 2 + 1:

17

460

changes to

460 WS=WN$

470

changes to 470 RETURN

服時546

務間:546

:每日上午十晚至下午五時半

(vi) Scanner

(6) The advantage. of VDU over the dot-matrix printer is that the

response time of VOU is much faster than the printer.

The disadvantage of VDU is that the image on it cannot be kept as hardcopy record.

(c) (i)

Keyboard can be used as an input device for the blind man because he can type by the position of the keys, a Braille Code Printer can be used as an output device because he can know the output by touching the raised dots

(1) A microphone can be used as input device for the limbless man because he can input by sound: A VDU can be used as output device because he can see the output: by eyes.

(d) Peripheral devices are needed because.

(i)

different applications,

tie devices can

used for

different

(11) The computer system may be cheaper by not using

expensive peripherals.

(iii) Some peripherals may be changed when it is damaged without discarding the whole expensive Computer System.

(a) Statement 1: get Christian name of a classmate

Statement 4.1: process Christian name. Statement 4,2: search list for a Christian name Statement 4.3Y: add the frequency of the Christian name to

the cumulative frequency.

Statement 4.3N: add the frequency of the Christian name to

the cumulative frequency. Statement 5 alphabetize the list.

START:

INPUT

NAMES

Execute Module B2

Assign 1 to

the frequency

of the Name

(b) (i)

The bit pattern is -17.

The ASCII code of "E" ts 69.

=1000101,

6910

The bit pattern is 1000101.

(14)

510

1012

000101 (6 bits)

= 1110102 (one's complement)

111011, (two's complement)

The bit pattern is 0111011.

(c) The smallest integer that can be represented is -32.

(d) the ASCII code of "1" is 49 which is equivalent to the denary

number 0110001

The ASCII code of "?" is 63 which is equivalent to the: denary number 0111111.

2. (a) 20 DIM R$111,6}-

・(b) OR

(c) IF LEN(CODE)<> 1 OR CODES < "O" OR CODES > THEN

(d) 1100 X VAL (R$ (ROOMNUM, VAL(CODES) 3))

1110 X$STRS(X+ AMOUNT).

1120 R$(ROOMNUM, VAL (CODES) +3)=X$

(e) This step is used to calculate number of days that the

customer occupied the room.

(f) Total fee (number of days accupied)*(room rate)+(breakfast charge) (service charge)+(Jaundry charge)

(g) 1500 FOR J = 1 TO 6-

1510 R$(ROOMNUM, 3)- 1520

NEXT J

(h) box. 3: Open the file

box 1.7: Close the file

3.

(a)

11010011

"1" and "?" cannot be represented.

AND. 10001101

The type bit will then be 0 which can represent numbers only.

10000001

(e) The answer to (c) remains unchanged because the first bit

does not affect the magnitude of the number.

The smallest number is -32.

The answer to (d) is changed,

"" cannot be represented because the ASCII code of it is 93. The binary equivalent is 1011101..

The type bit is then 1 which represents an

ceger

(a)

8:G

9:0

10 B

11

-12 I

The total number of marks and the range of marks should

(Only the solution is required.).

(b) (1)

·B.. ·BITPOS

DUOFF.

ELAG

91

.0

1.

$

g

3

8

8

1

32

0

(ii) The purpose of this subroutine is to FLAG

(c) (1)

testing.

INDEX= 12

BITPOS

4

* R(INDEX) = 17

(ii) Line 160 is used to keep record the voted status of the

members

(iii) Each bit of the binary representation is used to kept the voted status of a member. Each byte can handle 8 bers. Therefore R(1000) is sufficient for 8000

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be validated.

Search the

(11)

Name of

dent and mark,

list for the

(c) (1)

Christian

пале

Position in the class.

(ii) Class of the students.

Is

the name

found?

NO

Execute Module B2

Add frequency to the cumulative frequency of the

'name'

110

20

· 130

140

150.

(b) 500

11 names

inputted?

Alphabetize the Fist

510

520

530

Yes

540

550

.560

Out the

Name List

SUGGESTED ANSWER TO COMPUTER STUDIES, PAPER II

Use the following table to show the corrections required for the program to produce the above output..

570

580

590

600

610

620

Line Number

Correction(s) Required:

630.

STOP

10

changes to 10 DIM HOBBY$(3)

(c) 800

810

20

changes to

20. INPUT "ENTER NAME"; NAS

820

3. (a)

address

1011

address

1100

address 1111:

accumulator

50

changes to 60 PRINT "ENTER HOBBY" I;

830

(a) Data stored in computer system may be accessed easily by

the people data are stored in compu Computer is cheaper and

computer system.

poter azon is common

(b) (). A computer operator sells the data which he can access. (11) A student access his, marks in the computer system.

(111) The personal data of citizen can be accessed by

different departments easily.

(iv) The floppy disk storing the data is damaged. The data

on it cannot be accessed.

(c) (i)

All the user can access all the data by just entering. the system

access al

The secret code may not be secret because too manys people know it h

(11) Every user has a specific secret code so that the Secret code matches the specific user. Different user may access to different level of the data unauthorized access of data.

event.

S.

VOSUMO

COUNT = 12

320

ELECT = 0(1):

330

PRINT "CANDIDATE. CODE NO. OF VOTES"

340

FOR 1 TO 5 %.

*350

LA PRINT TAB(5); 1; TAB(20); C({}~

360

370

380

IF C(1) ELECT, THEN 390

COUNTY AND CONS

ELECT CE1).....

VOSUM = VOSUM + C(I)

400

NEXT IN

1

PRINT TAB(20);"

420

PRINT TAB(14) TOTAL VOSUM

490 PRINT SUR

1=1

PRINT "CANDIDATE #" COUNT; "IS THE NEWLY ELECTED

PRESIDENT."

READ TLINE${1},

IF TLINES(I) - "*" THEN 150

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1=T+E

GOTO 110

RETURN

COUNT = 1

FOR

4 TO 1-1

N$ : *.

KLENCTLINES(3))

FOR M 1.10 KUE

CH$ = MID$(TLINES(J), M, 1}) -

IF CH$<>"" AND MC>LEN{TLINES(I))

THEN WS WS+ CHS GOTO 610-

IF M = LEN(ILINES(J)) THEN W$

WORD$(COUNT) = W$*

COUNT COUNT + 1

HS+ CHS

W$=!

NEXT M

NEXT J

RETURN

INPUT "ENTER LEFT MARGIN (130) LM

TF LM≤1 OR LM) 30 THEN PRINT "ILLEGAL LEFT MARGIN, RE- ENTER" GOTO 800

MAT

INPUT "ENTER RIGHT MARGIN (50 - 80)"; RM

IF RM350 OR RM) 80 THEN PRINT ILLEGAL RIGHT MARGIN, RE

ENTER" GOTO 820

7114368

70

changes to 70 INPUT W$; GOSUB 390

840

RETURN

松柏I

1st pass

2

-9

2:

..9

(d) 1000

1=0

100

changes to 100 FLAGS "FALSE"

1010

1020

·LINMAX = RM - LM + 1

2nd pass

4

-B

6.

120

changes to

120 IF MNAME$="*". THEN 260

1030

PRINT

3rd pass.

6.

-7.

12

·1040

PRINT TAB(LM);

140

changes to 140 IF MSEX$-SEXS THEN 110

·1050

4th pass

8

-6

20

1060:

L+ LEN (WORDS())

1.50

changes to 150 IF (SEX$="M" OR SEX$="m") AND.

AGE MAGE THEN 110

· 1070

IF I>COUNT THEN 1120

5th pass

10

-5

30

1080

160

(b) (i).

The content of 1111 is 110.

(ii) The content of accumulator is 0.

(c). The program is used to sum up the first 20 even numbers:

changes to 160 IF (SEXS="F" OR SEXS="F"). AND

AGE MAGE THEN 110

1090

1100

1.

1110

GOTO 1040

185

Insert line 185. NEXT 1

1120

RETURN

IF LILINMAX THEN PRINT WORD$(I); "::

LL GOTO 1050

IF LLINMAX THEN PRINT WORD$(1): 10 GOTO 1040

*2*

12,

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