浙江选考信息技术客观题专题汇编之专题八树、队列、栈概念及应用（PDF版，答含答案）

1. 2022 5 8
1.
A. B. C. D.
2. 2022 6 10
2. a,c,e,h,r,t,e
A. c,h,e,a,t,e,r B. h,e,c,t,a,r,e C. t,e,a,c,h,e,r D. r,e,t,e,a,c,h
3. 2022 6 8
3 python 17 8
a,b = map(int,input().split( ))
w = 0
while b**w <= a:
w += 1
st = [0]*w ; top = -1
while a>0:
top = top+1
st[top] = a%b
a = a//b
while top > -1:
top = top-1
A 21 B 12 C 17 D 71
4. 2022 8 A9 9
4. 1,2,3,4,5
A 1,2,3,4,5 B 4,5,3,2,1 C 4,3,5,1,2 D 3,2,1,5,4
5. 2022 8 Z20 10
5 “ ” n
Python
top=-
while iwhile (1) and i(2)
st[top]=s[i]
i+=1
while top!=-1:
m+=st[top]
(3)
(1)(2)(3)
top-=1 top+=1 top+12／11
A B C D
6. 2022 9 6
6.
3 3
x1 x2 x3 x4 x5 x6
A. x1 x4 x6 x2 x3 x5 B. x4 x1 x6 x2 x3 x5
C. x1 x2 x3 x6 x4 x5 D. x6 x5 x4 x3 x1 x2
7. 2022 9 9
7.
A.
B.
C.
D.
8. 2022 9 10
8. head tail
head tail
que P y t h o n
0 1 2 3 4 5 6 ...
A. Python B. Ptoynh C. yhntPo D. yhntoP
9. 2022 9 11
9. Python ( :
“12321” “1232” )
def pal(s): #
* 100
top = -1
k= len(s) // 2
for i in range(k):
top += 1
st[top] = s[i]
if len(s) %2= 1:
1
for i in range(k, len(s)):
tmp = st[top]
top -= 1
if (2) :
return False
3／11
return True
k=k+1 k=k – 1 tmp== s[i] tmp != s[i]
(1) (2)
A. B. C. D.
10. 2022 10 9
10. 3
A.1 B.2 C.3
D.5
11. 2022 11 12
11. Python
a=[2,4,5,10,8,13,11,7,2,6]
que = [0] * len(a)
k = int(input())
key = int(input())
msq = 0;sq = 0
head = 0;tail = 0
for i in a:
que[tail] = i
sq = sq + i
tail = tail + 1
while sq > key or tail - head >= k:
sq = sq - que[head]
head = head + 1
if sq > msq:
msq = sq
k 3 key 20 msq
A. 18 B. 19 C. 20 D. 32
12. 2022 11 9
12. 1,2,3,4,5,6,7 1,4,3,2,7,6,5
A. 3 B. 4 C. 5 D.
6
13. 2022 9 8
13. head tail
head=2 tail=7
A. 2 B .5 C. 7 D. 9
14. 2022 9 9
14. :
4／11
s = [0] * 10;a = [6 3 2 4 2 1 5]
n = len(a);top = 0;s[top] = a[0]
for i in range(1, n):
while top != -1 and a[i] % 2 == s[top] % 2:
top -= 1
top += 1
s[top] = a[i]
while top != -1:
print(s[top] end = ' ')
top -= 1
.
A. 2 4 2 6 B. 5 1 3 C. 1 2 3 6 D. 5 2 3 6
15. 2023 10 7
15. Python
for i in range(len(s)):
q[tail]=s[i]
tail+=1
while headans=ans+q[head]
head+=1
if headq[tail]=q[head]
tail+=1
head+=1
print(ans)
s “QUEUE”
A. QUEUE B. EUEUQ C. QEUUE D. QEEUU
16. 2022 9 9
16. R1 R2 R3 R R5 R6 S Q
R4
A. R6 R5 R4 R3 R2 R1 B. R4 R6 R5 R1 R2 R3
C. R4 R5 R6 R3 R2 R1 D. R4 R5 R6 R2 R1 R3
17. 2022 10 11
17. n 1 n 1 m m>1)
Python
que=[0]*1000
n=int(input())
m=int(input())
tail,head=0,0
for i in range(1,n+1):
5／11
que[tail]=i;tail+=1
num=0
while head___________
head+=1;num+=1
if num==m:
____________
num=0
else:
___________
tail+=1
x=que[head] que[tail]=x
A. B. C. D.
18. 2022 11 12
18. Python
import random
n=len(s1);i=0
while iif random.randint(0,1)==0:
if len(s2)==0 or s1[i]>s2[-1]:
s2.append(s1[i])
i+=1
else:
if len(s2)>0:
s3+=s2.pop(-1)
while len(s2)>0:
s3+=s2.pop(-1)
print(s3)
A. saAsm B. sAise C. sAiaGmsea D. AsiaGames
19. 2022 9 9
19. a1, a2, a3, a4,
A. a2, a4, a3, a1 B. a3, a1, a4, a2 C. a3, a4, a1, a2 D. a4, a3, a1, a2
20. 2022 6 9
20 A-B-D-C-E D-B-E-C-A
A B-D-A-E-C
B B-D-A-E-C
C D-C-B-A-E
D D-C-B-A-E
21. 2022 8 Z20 11
21. BIGDHAECF IGHDBEFCA
6／11
A ABCDEFGHI B ABDGHICEF C ABDHGICEF
D ABDGIHCEF
22. 2022 9 10
22. T2 G-D-H-E-B-I-F-C-A
D-G-B-E-H-A-C-I-F T2
A. A-B-D-G-E-H-C-I-F B. A-B-D-G-E-H-C-F-I
C. A-B-D-G-E-H-F-C-I D.
23. 2022 9 10
23. CBADE BACED
A. ABEDC B. CBEDA C. ABDEC D. CBDEA
24. 2022 10 12
24. Python
import random
key=random.randint(0,4)*2+5
n=10;ans=0
a=[4,5,5,8,9,11,11,13,15,17]
i=0;j=n-1
while i<=j:
m=(i+j)//2
if a[m]<=key:
i=m+1
else:
j=m-1
ans+=a[m]
print(ans)
ans
A. 19 B. 27 C. 37 D. 44
25. 2022 10 11 +
25.
* 3
- 5
9 /
9-4/2 *5+3
4 2
A.+*-9/4253 B.+*-/42953 C.942/-*53+ D.942/-5*3+ 25
26. 2022 11 11
26. 12
7／11
A. 12 B. 3
C. 5 D. 5
27. 2022 11 9
27. A-G
27
0 1 2 3 4 5 6 7 8 9 10 11 12
A B C D E F G
27
A. DBAGECF B. BDAFECG C. ABDCEFG D. DBGEFCA
28. 2022 11 9
28. 3/(5*2)
A.
B. 2 /
C. 352*/
D. 3 *
/ 3 * 5 2 5 2
28
29. 2022 11 10
29. 29
29
A. 4
B. 25 15 46 6 18 28 58 12 22 35 60
C. 8
D. 0 “46” 2
30. 2022 11 12
30.
A. B.
C. D.
31. 2022 11 9
8／11
31. ABCDEF BCADFE
A. CBFEDA B. BCFEDA C. CBDFEA D. BCEFDA
32. 2022 7
32.
A. 3 14
B. 8 26
C. 14 8 3 12 26 15 27 28 3 12 15 27
D. 28 8
28
33. 2023 10 2
33. “ ”
A. B. C. D.
34. 2022 10
34. “abdgehicfj” “dgbheiafjc”,
A. dgjoebkfca B.gdhiebjfca C. gdbhiejfca D.gdhiefjca
35. 2022 6 10
35. 35
A.XBCDAYEF B.FEYADCBX
C.DBEAFXCY D.DEFABYCX
36. 2022 11 11
36.
35
A.
B.
C.
D.
37. 2022 11 10
37. */a+bc*+-defg
a/b+c*d-e+f*g
A. abc+/de-f+g** B. ab/cd*+e-fg+* C. abcd*+/e-+fg* D.
abcd*+e-/fg+*
38. 2022 10
38. ABCDEF, CBAEDF,
A. FEDCBA B. CBEFDA C. CBEDFA D.
39. 2022 12 9
9／11
39. :A-B-D-C-E :D-B-E-C-A
A. B-D-A-E-C
B. B-D-A-E-C
C. D-C-B-A-E
D. D-C-B-A-E
40. 2022 12 8
40. A B C D E
5
A. ABCED B. DBCEA C. CDABE D. DCBEA
41. 2022 12 11
41. Python T ['A' ,1,2]
'A' 1 2 ['K' ,-1,-1]
T HDIEBJKFGCA
:
T=[['A' ,1,2],['B' ,3,4],['C' ,5,6],['D' ,-1,7],['E' ,-1,8],['F' ,9,10],['G' ,-1,-1],['H' ,-1,-1],['I' ,-1,-1],['
J' ,-1,-1],['K' ,-1,-1]]
rsult=[]
def Lrd(node):
if T[node][1]!=-1:
Lrd(T[node][1])
elif T[node][2]!=-1:
Lrd(T[node][2])
rsult.append(T[node][0])
Lrd(0)
rsult
A. ['A', 'B', 'D'] B. ['H', 'D', 'B', 'A']
C. ['A', 'B', 'D', 'H'] D. ['D', 'B', 'A']
42. 2022 12 8
42. 60
, head, tail
(tail+1)%60==head head 54 tail 3 8
20 head tail
A. 1 22 B. 2 23 C. 23 2 D. 22 1
43. 2022 12 10
43. Python
import random
n=len(a);c=random.randint(0,n-1)
10／11
st=[0]*n;top=0;st[top]=0
for i in range(1,n):
while top != -1 and a[i]top -= 1;c -= 1
top+=1
st[top]=i
if c == 0:break
s=a[i+1:]
while top != -1:
s=a[st[top]]+s
top -= 1
-c])
23641287
A. 628 B. 687 C. 21 D. 12
44. 2022 12 8
44.
A
B C
D E
F
44
A.
B. 3
C. DBEFCA
D.
45. 2022 12 9
45. Python :
que=[0]*20
key=2
head=0;tail=0
for i in range(len(st)):
que[tail]=(int(st[i])+key)%10
tail+=1
else:
head+=1
while head!=tail:
print(que[head], end=' ')
head+=1
程序运行后，输出的结果是
A. 0 7 1 B. 5 0 7 1 C. 0 7 c 1 D. "5" "0" "7" "c" "1"
11／11
8. 树、队列、栈概念及应用
1 2 3 4 5 6 7 8
B A A C C B D C
9 10 11 12 13 14 15 16
D A A A B D D C
17 18 19 20 21 22 23 24
A B A B D B A A
25 26 27 28 29 30 31 32
A D A D D B A C
33 34 35 36 37 38 39 40
D B D D A B B D
41 42 43 44 45
B B D D A