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计算机2016年南京市江苏对口单招第一次调研试卷

计算机2016年南京市江苏对口单招第一次调研试卷
计算机2016年南京市江苏对口单招第一次调研试卷

南京市职业学校2013级对口单招第一次调研性统测

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一、单项选择题(本大题共30小题,每小题2分,共60分)

1 2 3 4 5 6 7 8 9 10

B B D D D A A B

C D

11 12 13 14 15 16 17 18 19 20

C B

D B C A C D D A

21 22 23 24 25 26 27 28 29 30

D C A D C D C B C A

二、判断题(本大题共15小题,每小题2分,共30分)

31 32 33 34 35 36 37 38 39 40

B B A B A B B B A B

41 42 43 44 45

A B B A B

三、填空题(本大题共30小题,每小题2分,共60分)

46.45 47.11000000 48.中断响应49.1

50.67FF 51.操作码52.时间控制53.微程序控制器

54.分时共享55.F10 56.NTFS 57.采样频率

58.240 59.PPM 60.单碟容量61.计算机网络

62.包过滤63.语义64.节点65.调制解调器

66.网络接口(访问)层67.网络地址68.CERNET 69.配置管理

70.25 71.3 72.2V 73.2

74.0 75.-2.68

四、案例分析题(本大题共2小题,每小题2分,共30分)

76.

(1)显卡———3分(2)ATI———3分

(3)28nm———3分(4)PCI-E——3分

(5)2 ————3分

77.

(1)CABLE(射频)———3分(2)光纤——————2分

(3)同轴电缆—————2分(4)双绞线—————2分

(5)级联——————3分(6)直通——————3分

78.(12分,每项4分,格式错,扣2分)

i=16,n =5

s=25

79.(12分,每项3分,格式错,扣2分)

x=22,y=-5

x=44,y=-10

80.(12分,每项3分)

①n=-n ②2

③r==0 ④n=n/i或n/=i

81.(12分,每项3分)

①return a ②t=t+a%10*p;

③num[i][0],num[i][1] ④fclose(fp)

82.程序设计,酌情判分。(12分,每个子函数6分)(1)

void findmax(int a[][N],int n[]){

int i,j,k;

for(i=0;i

k=0;

for(j=1;j

if(a[i][j]>a[i][k])k=j;

n[i]=k;

}

}

(2)

void move(int a[][N],int n[]){

int i,j,k,t;

for(i=0;i

k=n[i];

while(k!=0){

t=a[i][0];

for(j=1;j

a[i][j-1]=a[i][j];

a[i][3]=t;

k--;

}

}

}

83.解:

答83(b )图 答83(c )图

(1)当12V 电压源单独作用时,画出等效电路如答83图(b )所示:(1分) U ’= -10V (2分)

当3A 电流源单独作用时,画出等效电路如答83图(c )所示:(1分) U ’’= 4V (2分)

∴12V 电压源和3A 电流源共同作用时,U = U ’+U ’’ = -6V (2分) (2)3A 电流源的功率为 18W (2分)

(3)若要求3A 恒流源功率为零,则12V 恒压源应改为 4.8V (2分)

84.解:据题意,移去待求短路线AB ,电路如下图所示:(1分)

U AB =

15

19393244

-?+??++

= 6V (3分) R AB = 3Ω (2分)

根据戴维南定理,原电路的等效电路如右图:(1分) (1)I AB = 2A (2分)

(2)二极管的状态为 导通 (1分) 通过二极管的电流I AB = 1.77A (2分)

85.解:

(1)当S 1闭合,S 2断开时,

+

12V

+

_U ’4Ω

3A 4Ω

2Ω +

_

U ’’4Ω

9A

-15V +

A B

+6V -

I AB A

B

①当v 2为负半周时,处于导通状态的二极管是 V 1、V 4 ;(2分) ②则此时相应的直流输出电压的平均值V o = 9 V 。(2分) (2)当S 1断开,S 2闭合时, 10 V ;(2分)

(3)当S 1断开,S 2断开时,

根据题85(b )图所给的已知波形,画出输出电压V o 的波形。(3分)

V O

2

2V 0

2

2V -t

ω

(4)若S 1闭合,S 2闭合时, 14.14 V ;(3分) 86.解:

J

K

Q n+10001101

1

Q n Q n

01

CP

Q 0Q 1Q 2

Y

(a) JK 触发器的真值表 (b) 波形图 答86图

87.解:该电路逻辑表达式为Y =ABC C AB C B A BC A ???(3分) 化简后,Y =AC BC AB ++(2分) 真值表如下:(2分)

由真值表可知,该电路的逻辑功能是 少数服从多数 ;(2分)

用基本逻辑门实现如下(3分):

1

&

&&

Y

A

B

C

A B C Y 0 0 0 0 0 0 1 0 0 1 0 0 0 1 1 1 1 0 0 0 1 0 1 1 1 1 0 1 1

1

1

1

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