Reading Question - web.fg.tp.edu.tw
12
臺北市立第一女子高級中學 98 學年度 資訊學科能力競賽初賽 Reading & Question June 6, 2009 測驗時間:2009 年 6 月 6 日上午 08:10—09:00
Transcript of Reading Question - web.fg.tp.edu.tw
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1
PART I
n
10 3
2(1110)2 10
1 1 1 0
23 22 21 20 3 2 1 01 2 1 2 1 2 0 2 14× + × + × + × =
()
2
n
n n1 1
2+00
2
1. 2
2. bytes
4. 1+1
2 bit
bit 1
14 00001110 11110001 11110010
2
2. (1021) 3 10A: ( )10
3. (748) 2 bytes 2A:__________________
4. 2(10111011) 2 10 A:______________
3
PART II
X’ X
True True
False False
X0 = 0
X1 = X
XX = X
XX’= 0 X + 0 = X X + 1 = 1 X + X = X X + X’= 1
(1) a. X+Y = Y+X b. XY = YX
(2)
(3)
c. (W+X)(Y+Z)WY+WZ+XY+XZ
(4)
= (A + A') +B ()
Questions
2. Y =ABC+AB'C+ABC'+AB'C'
3. Y =A(B+C)A'+D
4. Y =AB+A'B
5. (Truth table)
T True F False
P Q ~P P∧Q P∨Q P⊕Q
T T F T T F
T F F F T T
F T T F T T
F F T F F F
A: _________ (A) Y ∨ (X ⊕ Y ⊕ Z)
(B) Y ∨ (X ∧ ~ Z) ∨ (~ X ∧ Z) (C) (X ⊕ Z) ∨ (Y ∧ ~ Z) ∨ (Y ∧ Z) (D) (X ⊕ ~ Y) ∨ (Y ⊕ ~ Z)
5
PART III
ADSLADSL Asymmetric Digital Subscriber Line
ADSL
12M/1M byte 1.5MB/128KB
. 203.64.52.1 8
255
IP ABC
(0)2 7()
2. B(128~191.x.x.x)
(10)2 14(+)
3. C(192~223.x.x.x)
(110)2 21
ADSL (A) (B)
(C)7.5 (D)
4. B? A: __________ (A) 120.9.2.1 (B)128.256.2.1 (C)192.168.0.1 (D)191.255.220.9
push poppushpop
enqueue dequeueEnqueuedequeue
Questions
1. Pop Push A Push B Push O Pop Pop Push AB Push Q Pop Pop Pop
2. Dequeue Enqueue A Enqueue B Enqueue C Enqueue D Dequeue Dequeue Dequeue Enqueue E Dequeue Dequeue
3. 3 1 2 5 4 6 Push
Pop Push Pop 1 2 3 4 5 6
10
PART V
Questions
1. int i, n = 0; char str[] = “6061”; for(i=0; ‘0’<=str[i]&& str[i]<=‘9’; i++) { n = n * 10 + (str[i] – ‘0’); } printf(“%d\n”, n);
2. if 1 100 2
3 6 for(i=1; i<=100; i++) { if( ) { printf(“%d\n”, i); } }
3. #include <stdio.h> #define SQUARE(X) X * X int main(void) { int n = 5; printf(“%d %d\n”, SQUARE (n), SQUARE (n + 1)); return 0; }
4. B Knm
n 5m 3(1) B(5,3)
____ ____ (2) ? ____ ____
Function B(n,m)
else K = B(n-1,m) + B(n-1,m-1);
PART I
n
10 3
2(1110)2 10
1 1 1 0
23 22 21 20 3 2 1 01 2 1 2 1 2 0 2 14× + × + × + × =
()
2
n
n n1 1
2+00
2
1. 2
2. bytes
4. 1+1
2 bit
bit 1
14 00001110 11110001 11110010
2
2. (1021) 3 10A: ( )10
3. (748) 2 bytes 2A:__________________
4. 2(10111011) 2 10 A:______________
3
PART II
X’ X
True True
False False
X0 = 0
X1 = X
XX = X
XX’= 0 X + 0 = X X + 1 = 1 X + X = X X + X’= 1
(1) a. X+Y = Y+X b. XY = YX
(2)
(3)
c. (W+X)(Y+Z)WY+WZ+XY+XZ
(4)
= (A + A') +B ()
Questions
2. Y =ABC+AB'C+ABC'+AB'C'
3. Y =A(B+C)A'+D
4. Y =AB+A'B
5. (Truth table)
T True F False
P Q ~P P∧Q P∨Q P⊕Q
T T F T T F
T F F F T T
F T T F T T
F F T F F F
A: _________ (A) Y ∨ (X ⊕ Y ⊕ Z)
(B) Y ∨ (X ∧ ~ Z) ∨ (~ X ∧ Z) (C) (X ⊕ Z) ∨ (Y ∧ ~ Z) ∨ (Y ∧ Z) (D) (X ⊕ ~ Y) ∨ (Y ⊕ ~ Z)
5
PART III
ADSLADSL Asymmetric Digital Subscriber Line
ADSL
12M/1M byte 1.5MB/128KB
. 203.64.52.1 8
255
IP ABC
(0)2 7()
2. B(128~191.x.x.x)
(10)2 14(+)
3. C(192~223.x.x.x)
(110)2 21
ADSL (A) (B)
(C)7.5 (D)
4. B? A: __________ (A) 120.9.2.1 (B)128.256.2.1 (C)192.168.0.1 (D)191.255.220.9
push poppushpop
enqueue dequeueEnqueuedequeue
Questions
1. Pop Push A Push B Push O Pop Pop Push AB Push Q Pop Pop Pop
2. Dequeue Enqueue A Enqueue B Enqueue C Enqueue D Dequeue Dequeue Dequeue Enqueue E Dequeue Dequeue
3. 3 1 2 5 4 6 Push
Pop Push Pop 1 2 3 4 5 6
10
PART V
Questions
1. int i, n = 0; char str[] = “6061”; for(i=0; ‘0’<=str[i]&& str[i]<=‘9’; i++) { n = n * 10 + (str[i] – ‘0’); } printf(“%d\n”, n);
2. if 1 100 2
3 6 for(i=1; i<=100; i++) { if( ) { printf(“%d\n”, i); } }
3. #include <stdio.h> #define SQUARE(X) X * X int main(void) { int n = 5; printf(“%d %d\n”, SQUARE (n), SQUARE (n + 1)); return 0; }
4. B Knm
n 5m 3(1) B(5,3)
____ ____ (2) ? ____ ____
Function B(n,m)
else K = B(n-1,m) + B(n-1,m-1);