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C新教材课后习题参考答案汇总

C新教材课后习题参考答案汇总
C新教材课后习题参考答案汇总

第一章一、判断题

第二章

三、填空题

1、1个

2、11.5

3、28 4 5

4、运算符的优先级结合性

5、double

6、int k1=7,k2=7; -2

7、2.5 7.5

8、2.000000

9、double型

10、ASCII码

第三章

二、阅读下列程序,写出运行结果

1、5678

11, 21

2、n=69 k=96

3、1 4 4

0 4 4

4、3 1 3 3 3 2

1 1 0 0 1 0

5、

(1)i的值为52,c的值为?$?,f的值为9.17,d1的值为3.141593 (2)i的值为52,c的值为?9?,f的值为0.17,d1的值为3.141593 (3)i的值为52,c的值为?9?,f的值为0.17,d1的值为3.141593 6、3.00%5.00%6.00 7.483

三、程序设计题

1、#define PI 3.1415

#include “stdio.h”

main()

{ float r,h,v;

printf(“input r,h:”);

scanf(“%f,%f”,&r,&h);

v=PI*r*r*h;

printf(“v:%f”,v);

}

2、#include “stdio.n”

main()

{ char a,b,c;

printf(“input a,b,c:”);

scanf(“%c,%c,%c”,&a,&b,&c);

printf(“%c,%d\n”,a,a);

printf(“%c,%d\n”,b,b);

printf(“%c,%d\n”,c,c);

}

3、#include “stdio.h”

main()

{

int k;

printf(“k=”);

scanf(“%d”,k);

printf(“%o,%x”,k,k);

}

4、#include “stdio.h”

main()

{

printf(“input ch:”);

scanf(“%c”,&ch);

clrscr();

printf(“%c %c\n”,ch,ch);

printf(“ %c %c\n”,ch,ch);

printf(“ %c %c\n”,ch,ch);

printf(“ %c\n”,ch,ch);

printf(“ %c %c\n”,ch,ch);

printf(“ %c %c\n”,ch,ch);

printf(“%c %c\n”,ch,ch);

}

5、#include “stdio.h”

main()

{

float cj1,cj2,cj3;

printf(“input cj1,cj2,cj3:”);

scanf(“%f,%f,%f”,&cj1,&cj2,&cj3);

printf(“总成绩为:%5.1f,平均成绩为:%4.1f”,c1+c2+c3,( c1+c2+c3)/3);

}

第四章

一、单项选择题

二、阅读下列程序,写出运行结果

1.F

2. Hello!

Good morning!

Bye-Bye!

3. x

三、程序填空题

1. _c>=?a?&&c<=?u?

__c=c-21_______

2. scanf(“%d,%d,%d”,&x,&y,&z);

y>z

四、程序设计题

1. #include

#include

main( )

printf(“input x”);

scanf(“%f”,&x);

if(x>0) printf(“%f”,sqrt(x));

else if(x<0) printf(“%f”,fabs(x));

else printf(“Bye,Bye”);

}

2. #include

main( )

{ char c;

c=getchar( );

if(c>=?A?&&c<=?Z?) putchar(c+32);

else c=getchar( );

}

3. #include

main()

{ int score,grade;

printf("Input score: ");

scanf("%d", &score);

grade= score / 10;

switch(grade)

{ case 0:

case 1:

case 2:

case 3:

case 4:

case 5: printf("grade=E"); break;

case 6: printf("grade=D"); break;

case 7: printf("grade=C"); break;

case 8: printf("grade=B"); break;

case 9:

case 10:printf("grade=A"); break;

default: printf("no score"); break;

}

}

4. #include

main()

{int x,sum;

scanf(“%d”,&x);

if(x>=1000&&x<=9999)

sum=x%10+x/10%10+x/100%10+x/1000;

printf(“sum=%d”,sum);

}

5.#include

void main()

{double rate,salary,tax;

scanf(“%lf”,&salary);

if(salary<850) rate=0;

else if(salary<1350) rate=0.05;

else if(salary<2850) rate=0.10;

else if(salary<5850) rate=0.15;

else rate=0.20;

tax=rate*(salary-850);

printf(“tax=%0.2lf\n”,tax);

}

6.#include

main()

{ char c;

c=putchar( );

if(c>=?a?&&c<=?z?) c=c+32;

else if(c>=?A?&&c<=?Z?) c=c-32;

printf(“%c”,c);

}

第五章一、单项选择题

二、阅读下列程序,写出运行结果

1、1 -2

2、5

3、1 3 5

4、a=8

5、a=3,i=7

6、1# 3# 21# 1173#

7、*#*#

三、程序填空题

1.fabs(item)>=0.00001

n++;

2.if(in<0)in=-in;

s=0;

digit=in%10;

in=in/10;

3. n!=0

n=n/10;

4. if(n<0)n=-n;或n=fabs(n);

s=s+n%10;

四、程序设计题

1.

#include"stdio.h"

main()

{int i;

long s=0;

for(i=1;i<=1000;i++)

if(i%2!=0) s+=i;

printf(“ld”,s);

}

2.

#include"stdio.h"

#include"math.h"

main ()

{

int n, m, i, flag,x,y;

scanf(“%d %d”,&x,&y);

if(x>y)

{ i=x; x=y;y=i;}

for(n=x;n<=y;n++)

{ m=sqrt(n);

flag=0;

for(i=2;i<=m;i++)

{if(n%i==0)

{flag=1;break;

}

}

if(flag==0)

printf("%5d",n);

}

}

3.

#include

#include

void main()

{ double s=0;

int i;

for(i=2; i<=10; i++)

s = s+sqrt(i);

printf( "%.10f\n", s);

}

4.

#include

void main()

{ long s=1, k=1,n;

int i;

scanf(“%ld”,n);

for(i=2; i<=n; i++)

{ k = k*i;

s = s+k;

}

printf("%ld", s);

}

5.

#include

main()

{float s,h;

int i;

s=100;h=s/2;

for(i=2;i<=10;i++)

{ s+=h*2;

h/=2;

}

printf(“s=%f\th=%f\n”,s,h);

}

6.

#include

void main()

{ int cock, hen, chick;

for(cock=0; cock<=20; cock++)

for(hen=0; hen<=35; hen++)

for(chick=0; chick<=100; chick+=3)

if(cock+hen+chick==100)

if(cock*5+hen*3+chick/3==100)

printf( "%d,%d,%d\n", cock, hen, chick);

}

7.

#include

#include

void main()

{ f loat s=0, a=81;

int i;

s = 81;

for(i=1; i<30; i++)

{a = sqrt(a);

s = s+a;

}

printf("%.3f", s);

}

第六章一、单项选择题

二、阅读下列程序,写出运行结果

1、1

2、1 3 4 5

3、hello

4、hello everybody

5、21

三、程序填空题

1、i==j

a[i][j]

2、&a[i]

i%4==0

putchar(…\n?);

3、str[0]

s

4、#include “math.h”

m%i==0

h%10==0

5、gets(s);

s[I]!=?c?

s[j]=?\0?;

四、编程题

1、

#include

main()

{int a[3][4],i,j,max;

for(i=0;i<3;i++)

for(j=0;j<4;j++)

scanf(“%d”,&a[i][j]);

max=a[0][0];

for(i=0;i<3;i++)

for(j=0;j<4;j++)

if (max

printf(“max=%d”,max);

}

2. main()

{char str1[80],str2[40];

int i=0,j=0;

printf("\n请输入字符串1:");

scanf("%s",str1);

printf("\n请输入字符串2:");

scanf("%s",str2);

while(str1[i]!='\0')

i++;

while(str2[j]!='\0')

str1[i++]=str2[j++];

str1[i]='\0';

printf("\n连接后字符串为%s",str1);

}

6. #include

#include

main()

{int a[30],i,j,t,n;

printf("请输入n的值:");

scanf("%d",&n);

printf("请输入%d个值:",n);

for(i=0;i

{scanf("%d",&a[i]);

if(a[i]<0) a[i]=fabs(a[i]);}

for(i=n;i>0;i--)

for(j=0;j

if(a[j]

printf("%4d",a[i]);}

7. #include

main()

{int f[11],i;

f[1]=0;

f[2]=1;

for(i=3;i<=10;i++)

f[i]=f[i-2]+f[i-1];

for(i=1;i<=10;i++)

{if((i-1)%5==0)

printf("\n");

printf("%5d",f[i]);

}

}

8. #include

#include

main()

{ int i,m;

char c[80];

scanf("%s",c);

m=strlen(c);

for(i=0;i<=m;i++)

{ if(c[i]>='a'&&c[i]<='u'||c[i]>='A'&&c[i]<='U')

c[i]=c[i]+5;

else

if(c[i]>='v'&&c[i]<='z'||c[i]>='V'&&c[i]<='Z')

c[i]=c[i]-21;

}

puts(c);

}

9. #include

#define N 100

main()

{char str[N],nstr[N],ch;

int i=0,j=0;

printf("文字:");

gets(str);

printf("字符");

scanf("%c",&ch);

while(str[i]!='\0')

{if(str[i]!=ch)

{ nstr[j]=str[i];j++;}

i++;}

nstr[j]='\0';

printf("字符串:%s\n",nstr);

}

第七章一、单项选择题

二、阅读下列程序,写出运行结果

1、7

2、1!=1

2!=2

3!=6

3、 6

4、-5 2 3 7 9

5、a=1

b=1

a=1

b=2

6、A+B=9

7、 6

8、6,5

9、0,3

10、-2

三、编程题

1.int isprime(int x)

{int m,i,flag;;

m=sqrt(x);

flag=0;

for(i=2;i<=m,i++)

{if(x%i==0) {flag=1; break;}}

if(flag==1) return 0;

else return 1;}

2.解法一:

#include

#include

int tgs(int x)

{int w;

w=(int)log10(x)+1;

if(x*x%(int)pow(10,w)==x) return 1; else return 0;

}

main()

{int i,w=0;

for(i=2;i<=10;i++)

if(tgs(i)) {w++;printf(“%4d”,i);}

}

解法二:

#include

int fun(int x)

{ if(x<10)

return !((x*x-x)%10);

else return! ((x*x*-x)%100);}

main()

{int x,y;

printf("please enter a interger number:"); scanf("%d",&x);

if(x>100 || x<0)

{printf("data ennor!\n");

exit(0);}

y=fun(x);

if(y) printf("%d yes\n",x);

else printf("%d no\n",x);}

3. 方法一:

lcd(int u, int v)

{int a, b, t, r;

if(u>v)

{t=u;u=v;v=t;}

a=u;b=v;

while((r=b%a)!=0)

{b=a;a=r;}

return(u*v/a);}

main()

{int u, v, h, l;

scanf("%d, %d",&u,&v);

printf("最小公倍数=%d\n",l);}

方法二:

hcf(u,v) /*最大公约数*/

int u,v;

{int a, b, t, r;

if(u>v)

{t=u;u=v;v=t;}

a=u;b=v;

while((r=b%a)!=0)

{b=a;a=r;}

return (a);}

lcd(u,v,h)

int u, v, h;

{return(u*v/h);}

main()

{int u, v, h, l;

scanf("%d, %d",&u,&v);

h=hcf(u,v);

printf("最小公倍数=%d\n",h); l=lcd(u,v,h);

printf("最大公约数=%d\n",l);} 6.方法一(递归法):

#include

void itos(int n)

{int i;

if((i=n/10)!=0)

itos(i);

putchar(n%10+?0?);

main()

{int number;

printf(“\n输入整数:”); scanf(“%d”,&number);

if(number<0)

{putchar(…-?);

number=-number; }

itos(number);

}

方法二(非递归法):

string(int n)

{char str[10];

int i;

{putchar(…-?);

n=-n; }

i=0;

do

{str[i++]=n%10+?0?;

n=n/10;}while(n>0);

while(--i>=0)

putchar(str[i]);}

9. #include

double mp(double x,double y)

{double rs=1;

while(y--!=0)

rs=rs*x;

return(rs);}

double mpp(double x,double y)

{if(y==1)

return x;

else return x*mpp(x,y-1);

}

main()

{ double x; int n;

scanf("%lf,%d",&x,&n);

printf("%f\n",mp(x,n));

printf("%f\n",mpp(x,n));

}

11. #include

#define MAX 1000

main()

{int c,i,flag,flag1;

char t[MAX];

i=0;flag=0;flag1=1;

printf(“\n输入十六进制数”);

while((c=getchar())!=?\0?&&i

{if(c>=?0?&&c<=?9?||c>=?a?&&c<=?f?||c>=?A?&&C<=?F?) {flag=1;

t[i++]=c;}

else if(flag) {t[i]=?\0?;printf(“\n十进制数%d\n”,htoi(t));}}} htoi(char s[ ])

{int i,n;

n=0;

for(i=0;s[i]!=?\0?;i++)

{if(s[i]>=?0?&&s[i]<=?9?) n=n*10+s[i]-…0?;

if(s[i]>=?a?&&s[i]<=?f?) n=n*16+s[i]-…a?+10;

if(s[i]>=?A?&&s[i]<=?F?) n=n*16+s[i]-…A?+10;}

return(n);}

12.递归方法:

#include "stdio.h"

main()

{

int n;

int fact();

scanf(“%d”,&n)

printf("\n:%d!=%d\n",n,fact(n));

}

int fact(j)

int j;

{

int sum;

if(j==0)

sum=1;

else

sum=j*fact(j-1);

return sum;

}

第八章一、单项选择题

二、程序设计题

1、

#define mod(i,j) (i)%(j)

main()

{

int x,y,z;

printf("imput two integer:\n");

scanf("%d%d",&x,&y);

z=mod(x,y);

printf("%d\n",z);

}

2、

#define l ower(i) (i)>=?a?&&(i)<=?z??(i)-32:i

main()

{

char x,y;

printf("imput a lowercase:\n");

scanf("%c",&x);

y=lower(x);

printf("%c\n",y);

}

第九章一、单项选择题

二、阅读下列程序,写出运行结果

1、HELLO!

2、ef

3、3,5,5,3

4、-9

5、CDG

三、程序填空题

1、num=*b

num=*c

2、*(a+i)=*(a+j)

*(a+j)

3、char *str

*p!=?\0?

*str=?\0?;

4、char *s

*s-…0?

n*flag

5、scanf(“%d”,&n);

n==7?weekname[0]:weekname[n]

四、程序设计题

1、

main()

{

int *a,*b,*c,num;

printf("输入3个数:");

scanf("%d%d%d",a,b,c);

if (*a<*b) {num=*a;*a=*b;*b=num;}

if (*a<*c) {num=*a;*a=*c;*c=num;}

if (*b<*c) {num=*b;*b=*c;*c=num;}

printf("%d,%d,%d\n",*a,*b,*c);

getch();

}

2、

#include

main()

{

int i,m,n,k,mx,nx,*a[10],*p=a;

for(i=0;i<10;i++,p++)

scanf("%d",p);

p=a;

mx=nx=*p;

for(i=1,m=1;i<9;i++)

{

if(mx<*(p+i))

{

m=i;

mx=*(p+i);

}

}

for(i=1,n=1;i<9;i++)

{

if(nx>*(p+i))

{

n=i;

nx=*(p+i);

}

}

k=*(p+m);

*(p+m)=*(p+n);

*(p+n)=k;

printf("The array has been reverted:\n");

for(p=a;p

printf("%d ",*p);

getch();

}

#include "stdio.h"

#include "conio.h"

main()

{

char *c;

int letters=0,space=0,digit=0,others=0;

printf("please input some characters\n");

while((*c=getchar())!='\n')

{

if(*c>='a'&&*c<='z'||*c>='A'&&*c<='Z')

letters++;

else if(*c==' ')

space++;

else if(*c>='0'&&*c<='9')

digit++;

else

others++;

}

printf("all in all:char=%d space=%d digit=%d others=%d\n",letters, space,digit,others);

getch();

}

4、

#include "stdio.h"

#include "string.h"

main()

{

char *string,*i,*j;

int m,n;

printf("please input a string:\n");

gets(string);

m=strlen(string);

for(i=string,j=string+m-1;i<=string+(m-1)/2;i++,j--)

{

if (*i!=*j) break;

}

if (i

printf("%s不是回文",string);

else

printf("%s是回文",string);

getch();

5、

#include "stdio.h"

main()

{

char *str;

int i=0;

printf("please input a string:\n");

gets(str);

while(str[i]!='\0')

{ if(str[i]>='a'&&str[i]<='z')

str[i]=str[i]-32;

i++;}

printf("%s\n",str);

getch();

}

7、

#include "stdio.h"

#include "conio.h"

main()

{

char str[100],*p;

printf("please input a string:\n");

gets(str);

for(p=str;*p;p++)

{

if(*p>='a'&&*p<='z'||*p>='A'&&*p<='Z')

*p=*p+5;

}

printf("%s\n",str);

getch();

8、

#include "stdio.h"

main()

{

int x[3][4],i,j,m,*p[3];

for(i=0;i<3;i++) p[i]=x[i];

汇编语言课后习题解答

第1章基础知识 检测点1.1(第9页) (1)1个CPU的寻址能力为8KB,那么它的地址总线的宽度为13位。 (2)1KB的存储器有1024个存储单元,存储单元的编号从0到1023。 (3)1KB的存储器可以存储8192(2^13)个bit,1024个Byte。 (4)1GB是1073741824(2^30)个Byte、1MB是1048576(2^20)个Byte、1KB是1024(2^10)个Byte。 (5)8080、8088、80296、80386的地址总线宽度分别为16根、20根、24根、32根,则它们的寻址能力分别为: 64(KB)、1(MB)、16(MB)、4(GB)。 (6)8080、8088、8086、80286、80386的数据总线宽度分别为8根、8根、16根、16根、32根。则它们一次可以传送的数据为: 1(B)、1(B)、2(B)、2(B)、4(B)。 (7)从内存中读取1024字节的数据,8086至少要读512次,80386至少要读256次。 (8)在存储器中,数据和程序以二进制形式存放。

(1)1KB=1024B,8KB=1024B*8=2^N,N=13。 (2)存储器的容量是以字节为最小单位来计算的,1KB=1024B。 (3)8Bit=1Byte,1024Byte=1KB(1KB=1024B=1024B*8Bit)。 (4)1GB=1073741824B(即2^30)1MB=1048576B(即2^20)1KB=1024B(即2^10)。(5)一个CPU有N根地址线,则可以说这个CPU的地址总线的宽度为N。这样的CPU 最多可以寻找2的N次方个内存单元。(一个内存单元=1Byte)。 (6)8根数据总线一次可以传送8位二进制数据(即一个字节)。 (7)8086的数据总线宽度为16根(即一次传送的数据为2B)1024B/2B=512,同理1024B/4B=256。 (8)在存储器中指令和数据没有任何区别,都是二进制信息。

全新版大学英语综合教程2[第二版]课后答案解析

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