《c++程序设计》谭浩强课后习题答案.docx

上传人:b****5 文档编号:6805500 上传时间:2023-01-10 格式:DOCX 页数:179 大小:52.25KB
下载 相关 举报
《c++程序设计》谭浩强课后习题答案.docx_第1页
第1页 / 共179页
《c++程序设计》谭浩强课后习题答案.docx_第2页
第2页 / 共179页
《c++程序设计》谭浩强课后习题答案.docx_第3页
第3页 / 共179页
《c++程序设计》谭浩强课后习题答案.docx_第4页
第4页 / 共179页
《c++程序设计》谭浩强课后习题答案.docx_第5页
第5页 / 共179页
点击查看更多>>
下载资源
资源描述

《c++程序设计》谭浩强课后习题答案.docx

《《c++程序设计》谭浩强课后习题答案.docx》由会员分享,可在线阅读,更多相关《《c++程序设计》谭浩强课后习题答案.docx(179页珍藏版)》请在冰豆网上搜索。

《c++程序设计》谭浩强课后习题答案.docx

《c++程序设计》谭浩强课后习题答案

第一章

1.5题

#include

usingnamespacestd;

intmain()

{

cout<<"This"<<"is";

cout<<"a"<<"C++";

cout<<"program.";

return0;

1.6题

#include

usingnamespacestd;

intmain()

{

inta,b,c;

a=10;

b=23;

c=a+b;

cout<<"a+b=";

cout<

cout<

return0;

}

1.7七题

#include

usingnamespacestd;

intmain()

{

inta,b,c;

intf(intx,inty,intz);

cin>>a>>b>>c;

c=f(a,b,c);

cout<

return0;

}

intf(intx,inty,intz)

{

intm;

if(x

elsem=y;

if(z

return(m);

}

1.8题

#include

usingnamespacestd;

intmain()

{

inta,b,c;

cin>>a>>b;

c=a+b;

cout<<"a+b="<

return0;

}

1.9题

#include

usingnamespacestd;

intmain()

{

inta,b,c;

intadd(intx,inty);

cin>>a>>b;

c=add(a,b);

cout<<"a+b="<

return0;

}

intadd(intx,inty)

{intz;

z=x+y;

return(z);

}

2.3题

#include

usingnamespacestd;

intmain()

{charc1='a',c2='b',c3='c',c4='\101',c5='\116';

cout<

cout<<"\t\b"<

return0;

}

2.4题

#include

usingnamespacestd;

intmain()

{charc1='C',c2='+',c3='+';

cout<<"Isay:

\""<

cout<<"\t\t"<<"Hesays:

\"C++isveryinteresting!

\""<<'\n';

return0;

}

2.7题

#include

usingnamespacestd;

intmain()

{inti,j,m,n;

i=8;

j=10;

m=++i+j++;

n=(++i)+(++j)+m;

cout<

return0;

}

2.8题

#include

usingnamespacestd;

intmain()

{charc1='C',c2='h',c3='i',c4='n',c5='a';

c1+=4;

c2+=4;

c3+=4;

c4+=4;

c5+=4;

cout<<"passwordis:

"<

return0;

}

3.2题

#include

#include

usingnamespacestd;

intmain()

{floath,r,l,s,sq,vq,vz;

constfloatpi=3.1415926;

cout<<"pleaseenterr,h:

";

cin>>r>>h;

l=2*pi*r;

s=r*r*pi;

sq=4*pi*r*r;

vq=3.0/4.0*pi*r*r*r;

vz=pi*r*r*h;

cout<

:

fixed)<

:

right)

<

(2);

cout<<"l="<

cout<<"s="<

cout<<"sq="<

cout<<"vq="<

cout<<"vz="<

return0;

}

3.3题

#include

usingnamespacestd;

intmain()

{floatc,f;

cout<<"请输入一个华氏温度:

";

cin>>f;

c=(5.0/9.0)*(f-32);//注意5和9要用实型表示,否则5/9值为0

cout<<"摄氏温度为:

"<

return0;

};

3.4题

#include

usingnamespacestd;

intmain()

{charc1,c2;

cout<<"请输入两个字符c1,c2:

";

c1=getchar();//将输入的第一个字符赋给c1

c2=getchar();//将输入的第二个字符赋给c2

cout<<"用putchar函数输出结果为:

";

putchar(c1);

putchar(c2);

cout<

cout<<"用cout语句输出结果为:

";

cout<

return0;

}

3.4题另一解

#include

usingnamespacestd;

intmain()

{charc1,c2;

cout<<"请输入两个字符c1,c2:

";

c1=getchar();//将输入的第一个字符赋给c1

c2=getchar();//将输入的第二个字符赋给c2

cout<<"用putchar函数输出结果为:

";

putchar(c1);

putchar(44);

putchar(c2);

cout<

cout<<"用cout语句输出结果为:

";

cout<

return0;

}

3.5题

#include

usingnamespacestd;

intmain()

{charc1,c2;

inti1,i2;//定义为整型

cout<<"请输入两个整数i1,i2:

";

cin>>i1>>i2;

c1=i1;

c2=i2;

cout<<"按字符输出结果为:

"<

return0;

}

3.8题

#include

usingnamespacestd;

intmain()

{inta=3,b=4,c=5,x,y;

cout<<(a+b>c&&b==c)<

cout<<(a||b+c&&b-c)<

cout<<(!

(a>b)&&!

c||1)<

cout<<(!

(x=a)&&(y=b)&&0)<

cout<<(!

(a+b)+c-1&&b+c/2)<

return0;

}

3.9题

include

usingnamespacestd;

intmain()

{inta,b,c;

cout<<"pleaseenterthreeintegernumbers:

";

cin>>a>>b>>c;

if(a

if(b

cout<<"max="<

else

cout<<"max="<

elseif(a

cout<<"max="<

else

cout<<"max="<

cout<

return0;

}

3.9题另一解

#include

usingnamespacestd;

intmain()

{inta,b,c,temp,max;

cout<<"pleaseenterthreeintegernumbers:

";

cin>>a>>b>>c;

temp=(a>b)?

a:

b;/*将a和b中的大者存入temp中*/

max=(temp>c)?

temp:

c;/*将a和b中的大者与c比较,最大者存入max

*/

cout<<"max="<

return0;

}

3.10题

#include

usingnamespacestd;

intmain()

{intx,y;

cout<<"enterx:

";

cin>>x;

if(x<1)

{y=x;

cout<<"x="<

}

elseif(x<10)//1≤x<10

{y=2*x-1;

cout<<"x="<

}

else//x≥10

{y=3*x-11;

cout<<"x="<

}

cout<

return0;

}

3.11题

#include

usingnamespacestd;

intmain()

{floatscore;

chargrade;

cout<<"pleaseenterscoreofstudent:

";

cin>>score;

while(score>100||score<0)

{cout<<"dataerror,enterdataagain.";

cin>>score;

}

switch(int(score/10))

{case10:

case9:

grade='A';break;

case8:

grade='B';break;

case7:

grade='C';break;

case6:

grade='D';break;

default:

grade='E';

}

cout<<"scoreis"<

return0;

}

3.12题

#include

usingnamespacestd;

intmain()

{longintnum;

intindiv,ten,hundred,thousand,ten_thousand,place;

/*分别代表个位,十位,百位,千位,万位和位数

*/

cout<<"enteraninteger(0~99999):

";

cin>>num;

if(num>9999)

place=5;

elseif(num>999)

place=4;

elseif(num>99)

place=3;

elseif(num>9)

place=2;

elseplace=1;

cout<<"place="<

//计算各位数字

ten_thousand=num/10000;

thousand=(int)(num-ten_thousand*10000)/1000;

hundred=(int)(num-ten_thousand*10000-thousand*1000)/100;

ten=(int)(num-ten_thousand*10000-thousand*1000-hundred*100)/10;

indiv=(int)(num-ten_thousand*10000-thousand*1000-hundred*100-ten*10);

cout<<"originalorder:

";

switch(place)

{case

5:

cout<

dl;

cout<<"reverseorder:

";

cout<

break;

case4:

cout<

cout<<"reverseorder:

";

cout<

break;

case3:

cout<

cout<<"reverseorder:

";

cout<

break;

case2:

cout<

cout<<"reverseorder:

";

cout<

break;

case1:

cout<

cout<<"reverseorder:

";

cout<

break;

}

return0;

}

3.13题

#include

usingnamespacestd;

intmain()

{longi;//i为利润

floatbonus,bon1,bon2,bon4,bon6,bon10;

bon1=100000*0.1;//利润为10万元时的奖金

bon2=bon1+100000*0.075;//利润为20万元时的奖金

bon4=bon2+100000*0.05;//利润为40万元时的奖金

bon6=bon4+100000*0.03;//利润为60万元时的奖金

bon10=bon6+400000*0.015;//利润为100万元时的奖金

cout<<"enteri:

";

cin>>i;

if(i<=100000)

bonus=i*0.1;//利润在10万元以内按10%提成奖金

elseif(i<=200000)

bonus=bon1+(i-100000)*0.075;//利润在10万元至20万时的奖金

elseif(i<=400000)

bonus=bon2+(i-200000)*0.05;//利润在20万元至40万时的奖金

elseif(i<=600000)

bonus=bon4+(i-400000)*0.03;//利润在40万元至60万时的奖金

elseif(i<=1000000)

bonus=bon6+(i-600000)*0.015;//利润在60万元至100万时的奖金

else

bonus=bon10+(i-1000000)*0.01;//利润在100万元以上时的奖金

cout<<"bonus="<

return0;

}

3.13题另一解

#include

usingnamespacestd;

intmain()

{longi;

floatbonus,bon1,bon2,bon4,bon6,bon10;

intc;

bon1=100000*0.1;

bon2=bon1+100000*0.075;

bon4=bon2+200000*0.05;

bon6=bon4+200000*0.03;

bon10=bon6+400000*0.015;

cout<<"enteri:

";

cin>>i;

c=i/100000;

if(c>10)c=10;

switch(c)

{case0:

bonus=i*0.1;break;

case1:

bonus=bon1+(i-100000)*0.075;break;

case2:

case3:

bonus=bon2+(i-200000)*0.05;break;

case4:

case5:

bonus=bon4+(i-400000)*0.03;break;

case6:

case7:

case8:

case9:

bonus=bon6+(i-600000)*0.015;break;

case10:

bonus=bon10+(i-1000000)*0.01;

}

cout<<"bonus="<

return0;

}

3.14题

#include

usingnamespacestd;

intmain()

{intt,a,b,c,d;

cout<<"enterfournumbers:

";

cin>>a>>b>>c>>d;

cout<<"a="<

if(a>b)

{t=a;a=b;b=t;}

if(a>c)

{t=a;a=c;c=t;}

if(a>d)

{t=a;a=d;d=t;}

if(b>c)

{t=b;b=c;c=t;}

if(b>d)

{t=b;b=d;d=t;}

if(c>d)

{t=c;c=d;d=t;}

cout<<"thesortedsequence:

"<

cout<

return0;

}

3.15题

#include

usingnamespacestd;

intmain()

{intp,r,n,m,temp;

cout<<"pleaseentertwopositiveintegernumbersn,m:

";

cin>>n>>m;

if(n

{temp=n;

n=m;

m=temp;//把大数放在n中,小数放在m中

}

p=n*m;//先将n和m的乘积保存在p中,以便求最小公倍数时用

while(m!

=0)//求n和m的最大公约数

{r=n%m;

n=m;

m=r;

}

cout<<"HCF="<

cout<<"LCD="<

return0;

}

3.16题

#include

usingnamespacestd;

intmain()

{charc;

intletters=0,space=0,digit=0,other=0;

cout<<"enteroneline:

:

"<

while((c=getchar())!

='\n')

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

letters++;

elseif(c=='')

space++;

elseif(c>='0'&&c<='9')

digit++;

else

other++;

}

cout<<"letter:

"<

"<

"<

other:

"<

return0;

}

3.17题

#include

usingnamespacestd;

intmain()

{inta,n,i=1,sn=0,tn=0;

cout<<"a,n=:

";

cin>>a>>n;

while(i<=n)

{tn=tn+a;//赋值后的tn为i个a组成数的值

sn=sn+tn;//赋值后的sn为多项式前i项之和

a=a*10;

++i;

}

cout<<"a+aa+aaa+...="<

return0;

}

3.18题

#include

usingnamespacestd;

intmain()

{floats=0,t=1;

intn;

for(n=1;n<=20;n++)

{

t=t*n;//求n!

s=s+t;//将各项累加

}

cout<<"1!

+2!

+...+20!

="<

return0;

}

3.19题

#include

usingnamespacestd;

intmain()

{inti,j,k,n;

cout<<"narcissusnumbersare:

"<

for(n=100;n<1000;n++)

{i=n/100;

j=n/10-i*10;

k=n%10;

if(n==i*i*i+j*j*j+k*k*k)

cout<

}

cout<

return0;

}

3.20题

#include

usingnamespacestd;

intmain()

{constintm=1000;//定义寻找范围

intk1,k2,k3,k4,k5,k6,k7,k8,k9,k10;

inti,a,n,s;

for(a=2;a<=

展开阅读全文
相关资源
猜你喜欢
相关搜索
资源标签

当前位置:首页 > PPT模板 > 商务科技

copyright@ 2008-2022 冰豆网网站版权所有

经营许可证编号:鄂ICP备2022015515号-1