(31)育教 日七初月十年巴巴夏·
報日僑華
六期星 日四月一十九八九一)年八十七國民華中
日光
Discussions (1) The values stored in the array A(1), A(2),...A(11) Exercise on flowchart tracing:.
are covered..
"1990 中學再考預習連瀾
Computer (5)
MILL & DALE PRESS
K. Choy
LUIGA, MAKAN
Search to look for the existence of a particular number
character string in a sorted sequence.
Binary search method is a systematic way to search.
Algorithm: (a) Look for the middle term. If there are a terms in the sequence, the position of the mid-tera can be
calculated by
of start and end term
sure of the positions of
(b) Compare the selected middle term with the particular
number to be search,
(c) (i)
E
7.
=
20,
94.
Figures flowchart of binary search. Suppose there are (2) Steps (a), (b) and (c) are repeated until the and 10 elements in ascending order in an array.
of the list.
[R(1) 6. N(2) =
N(3) - 15, N(4) -17, N(5) = 24, M(7) 37, N(8) = 64, N(9) 72. N(10). - (3) At each round, half of the terms in the remaining (6) list are discarded because we are certain that the
-Complete the following number would not turn up in this half as the terms and X is the number to be looked for.
Lagles. (A round means the loop between boxes 5 and 10) are arranged in escending order.
(ii) Similar steps are used to locate the position of 160.
(A(1) M2) |A(3) A(4) A(5) A(6) A(7) A(6) A(9) A(10) |A(11)
(11 terms)
110
(i)
Number to be looked for : X. = 24
ROUND
I
Start.
N(1)
?
LOW
HIGH
10
LOWSHIGH!
2 3
(2 teras left) (i) Number to be looked for : X =-8
110 < 160
A(7) A(S) A(9) M(10) A(11)
(5 terms left)
159
what is the printed outout?
159 160
(10) A(11)
300 }
ROUND
I
300 160
(Not found)
the unsition if bota
Start
i
N(1)
!
1
A
elements in the list are discarded. in the list.
The number 150 is not
2
3
4
Discard the upper half (including the mid- term if the numbers loss than the mid-zerin (ii) Discard the lower half (including the mid- term) if the tomber 3 qreater than the mid- term.
(iii) The ters
is found
numbers are equal.
(0) The above three stops are repeated until the last
number remind is compared.
(e) If the number is not found for the whole search
process, the number is not in the array.
Example: Suppose theen is a sorted array of eleven numbers.
numbers storad are
1
START
The
2
the number to
be lacked few:
21, 35, 47, 99, 131, 10, 144, 145, 159, 300, 310. If the numbers are covered and we cannot see the values. Search for the existence of (1) 101 and (ii) 160.
Solution: (i)
3
LOW - 1
HIGH- 10
What is the printed output?
Unit
A
LOW
HIGH
LOWONIGH
Hyi
10
(ii) From the above example, to search a number in an array of 10
elencats, there are at most 4 trials (rounds).
What is the number of trials at most if an array contains 20 elements?
Position
Num
(a)
ויני
45 3:0
aware than at me in A151 with 101.
1- Inteyrsl
M:1 al
({LOWNISCH)/23
(c)
AS ONLY 170, the elements After A(5) as discarded,
f
X = N
YES
35
:M
[6]
(c)
1 ! -try n;"%།༢།་re
teen
3rd Teen
tommee the when shared to 4131 E 101.
As (0) 47, the plementy before 604) 16 esscarded.
A141 MSH
99
10
YI.S
LOW HIGI
NO
Answer: (i) Number to be looked for : X = 24
ROUND
1
N(1)
LOW
RIGH
|LOW HIGH}
Start
1.
1
19
20
む
10
2
64
6
3
24
CONY
/
Yes
Yes
The outgut is "The number is at 6th position.
I
(ii) Number to be looked For: X = 18
ROUND
N(I)
LOW
HIGH
LOWSHIGH
tart
10
20
?
Yes
15
A
Yes
17
5
4
No
AM
thi
1990 中學會考預蠶專欄
出版社
MIET & DALE PRESS
Mathematics (5)
12
the number
Vis NOT IN
-ray"
STOP
.....(3)
Subsitute (3) nto (2)
10. :-wwkeat post equations
A
NG KONG
HONG
3. | 4x
3
..(1)
!.....(2)
(4) - (3)
2b = 5
solutions, determine
From (1)
53
Solutions
Section A
1.
:.
Exercise 5:
Section A
H. K. LO
Simultaneous Equations.
(1)
1. Selve the stem of equations
1.
fir
!
2. valve the system and aggtropy
3. Sale the system of aquattusi
1
......(3)
bate (3) PL (2)
Ly
6x4-7x-13 - 1
40+13)-1) = 0
13
X = 0
1
Zx2
4x + 1 = 0
= 1 =0.7021
PUBLIC
When x =
1.7071
3 4130711 -0.4393
when x 0.2929
y = 470-29297 = 2.5506
= 1.7071
0.4393
orl*
0.2929
(Ans.)
= 2.5606
2 x (3)
28 +
(5)
1
:
The number is NOT in the array."
財
(iii) of an array contains 20 events. the number う
(rounds) is 5.
trials
全帽
RIES
8.
Let the distance be 5, speed.
v be the
港
即新
.(1)
3-
(Ans.)
(2) = (1)
25
w
-(2)
九龍旺
41
6. Let ax! 2. b = y + 3
1
1
=
5
.{1}
{Ans.)
v{v÷1)
3v(y-1) - Evlv-t)
For v 0
3(v+1)=5(v-1)
a+3b= 40 .....(2)
Froz (2)
a = 40- 3b .....(3)
Substitute v 4 into (1)
311-36
+
1
5
b-(49-35) = {{b}{40-35}
394 - 50b - 200 - 0.
5
4
廠
AR 126
化、適合省
?
x2 + 16y2 = 25 5x2-19xy-3y0.....(2)
.(1)
.. (b = 10)(35 - 20) = C
.*. b 10 Car
20
5 = 25
20
From (2)
When 16
4.
Wow the water!
2
(6x-y)(x-3)= 0
y = 6x
(3)
3y
.(4)
- 11/
5. Selve to waston of waratan
(ins, F
Consider y = 6x
.. y = 7
a = 10-35 = 10
3 = 10
1 x + 2 = 10
and
Substitute into (1)
x2 -16(6x)2 = 25
When b
9.
3x + y = ......{1}
x2 + y2 = 25 .....(2)
X = 8
(Ans.)
From (1)
y =
[Ans.)
the distance between A and B is 25 km.
尖沙咀星
3-735186
中医皇后:
5-217205
5-890713
旺角週
3-125655
荃灣地
0 49207
官嫵地鐵 3-797302 票: 釐期一
6. Selve me sistem of acetam
(2)
15...(1)
.(2)
2....(3)
Section B
7.
If the equations Bary Shell, Precedent ant (p-1)x-y 1:0 and all satesfond simulta musly is une pare of carui y. Pet the value
f
x'
577
x = 0.708 or
2.08
x = -0.208
- 1.249
་ ་
LY = -1.249
natirate (3) into (1)
Consider x 34
Substitute into (1)
(3)
16y= 25
16
y' = 1
101
J. 40
35 = 20
y + 3
नु
¡x - -0.203 (Ars.)
20
25x2
and x 12
[Ans.!
Since
it is
only ONE root
1 Cr
7-1
3
-3
(Ans.)
y=-1
Section B
7.
V - 5 = 0 3xy 49...
(1) .(2)
(1) + (2)
10.
ex - 100
..1.25
(K-3292 - 25
16x2 + (k-3x)2 = 400
16x2 + k2 - 6kx 9x2 - 400
6kx - (k2-400) = 0
a Langent, there is
‚*. (-6x)2 - 4(25)(k2-400) = 0
-64k2 + 40000 = 0
.*. k = 25 m k = -25
(Ans.)
t
» 馬豪馬七艹馬
یا
8.
A
A
un fan speed.
-3.23
.927
reduce, hy
speed by 1 kes par hrun the journey will te
hall's
www. good! be 1 km, your,
Substitute into (2)
-3.235
- 3 .....(1)
at will take 2
ניין:
i
more.
Pari qistance best woon: A
- 9 .....(2)
y = -0.25 Substitute the values at into
X
and y
and B.
9. 1 3x+4y-K
circle,
constant.
4,927
tanged af
1
whoes
201
BULQ2715 - 3,891
-2.353
ing.
Ans
Let a = and c =
a. 2b 3 .....(3)
a : 4b = 9 .....(4)
(p-1)x - 8y +1 = 0
.. (0-11.25)-8(-0.25):1-0
from which
D = -1.4
(Ans.)
1 A .........(2)
Substitute (2) into (1)
X + Ķ - x2 - 5x + 4
.*. * - 6x + (4-k1 = 0.....(3)
For no real solution,
B2 - 4AC < 0
(-6)2 - 4(1)(4~k) < 0
36 - 15+ 4K CO
4k < -20
k <-5
[Ans.)