c++程序设计谭浩强课后习题答案(完整版)，方便大家学习C++，课后答案很详细，有讲解，对于C++初学者来说是一份宝贵的资料！
return 0 27题 #include <iostream> using namespace std int main( Hint i,j,m, n; j=10: m=++|t++ n=(++)+(+j+m cout<<k<\t'<<j<<\t'<<m<<\t'<<n<<endl return o 2.8 #include <iostream> using namespace std; int main() (char c1=C, c2=h, c3=i, c4='n, c5=a C+ 3+=4 5+=4 cout<< password is: <<c1<<c2<<c3<<c4<<c5<<endl return 0 3.2题 #include <iostream> #include <iomanip> using namespace s td int main () oath, r, ,s, sq,vq, VZ: const float pi=3, 1415926 cout<<"please enter r, h: cin>>r>>h = pi s=reply sg=4 pirr vq=3.0/40·pi"r’r"r; Vz=pir"r"h cout<<setiosflags(ios: fixed)<<setiosflags(ios: right <<setprecision(2) cout<<l="<<setw(10)<<k<endl; cout<<"s=<<setw(10)<<s<<endl cout<< sq="<<setw(10)<<sq<<endl; cout<< vq="<<setw(10)<<vq<<endl; cout<< vz=<<setw(10<<vz<<endl; return 0 #include <iostream> using namespace std; int main 0 cout<"请输入一个华氏温度 din>冲t 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:"<<c<end return 0; 34题 #include <iostream> using namespace std; int main () Ichar c1, C2; cout<"请输入两个字符c1,c2 1=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<<endl cout<“"用cout语句输出结果为:"; cout<<c1<<c2<<endl return o 34题另一解 #include <iostream> using namespace std: int main Ichar c1, c2 cout<"请输入两个字符c1,c2"; getchar: ∥/将输入的第一个字符赋给c1 =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<<endl cout<<"用cout语句输出结果为:"; cout<<c1<<"<<c2<<endI; return 0. 5题 #include <iostream> using namespace std int main() char c1, c2; int ili. /定义为整型 cout<"请输入两个整数诅n: n>>i1>>2 c2=2 cout<"按字符输出结果为:<c1<","<<c2<<endl return 0; 38题 #include <iostream> using namespace std; int main f int a=3, b=4, c-5, x, y? cout<dla+b>c&& b==ck<<endl cout<<(a b+c&&b-ck<<endl; cout<<(!(a>b)&&Ic| 1)<<endl; cout<<(!(x=a)&&(y=b)&&0)<end cout<<((a+b)+c-1 && b+c/2)<<endl return O indlude <iostream> using namespace std; int main () t.b cout<< please enter three integer numbers: cin>>a>>b>>C: if(a<b) cout<< max Ise cout<<"max="<<b else if (a<c) cout<< max=<<c se cout<< max=<<a cout<<endl return 39题另一解 #include <iostream> using namespace std int main Hint a, b, c, temp, max cout<< please enter three integer numbers:", 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ma cout<< ma=<<max<send return 0 310题 #include <iostream> using namespace std int main () X,yr cout< enter. n>>X; if (x<1 -Xi cout<<"x="<<",y=x="<<y; else if (x<10 ∥/1≤x<10 cout<x="<<",y=2x-1="<<y; else ∥/x≥10 y=3*×-11; cout<"x="<xx<",y=3x-11="<y cout<<en return o 311题 #include <iostream> using namespace std; int main () float score: char grade cout<<"please enter score of student cin>>score while(score>100<O) Icout<< "data error, enter data again i cIn>>score switch(int(score/10) (case 10 case 9: grade='A break; case 8: grade=B: break case 7: grade =C- break: case 6: grade=D: break; default: grade=E: cout<< "score is"<<score<<, grade is <<grade<<endl return 0 3.12 题 #include <iostream> using namespace std, int main( Hong int num; int indiv, ten, hundred, thousand, ten thousand, place: /*分别代表个位,十位, 百位千位万位和位数 cout<<enter an integer(0-99999): cin>>num: if (num>9999) place=5 else if (num>999) place=4; else if(num> 99 place=3; else if(num>9) place=2; else place=1; cout<<"place="<<place<<endI; ∥)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-hund red100/10 indiv=(int)(num-ten thousand10000-thousand1000-hun dred"100-ten10): cout<< original order: switch(place) case 5: cout<<ten thousand<<, <<thousand<<,<<hundred< <,<<ten<< ,<<indiv<<en cout<< reverse order. cout<<indiv<<ten<<hundred<<thousand<<ten thousand< <end Case 4: cout<<thousand<<"<<hundred<<<<ten<<.<<indiv <<endl cout<< " reverse order. cout<<indiv<<ten<hundred<thousand<<endl break case 3: cout<<hundred<<"<<ten<<"<<indiv<<endl cout<< reverse order: cout<<indiv<<ten< <hundred<<endl eak: case 2: cout<<ten<<",<<indiv<<endl cout<< reverse order: cout<<indiv<<ten<<endl break case 1 cout<<indiv<<endl cout<< reverse order: cout<<indiv<<endl; break turn 0: 313题 #include <iostream> using namespace std int main f long i: 为利润 float bonus bon1, bon 2 bon 4 bon 6. bon 10- bon1=100000°01 /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+1000000.03 /利润为60万元时的奖 bon10=bon6+400000*0.015: /利润为100万元时的 奖金 cout<< enter i:i cIn>>I if(i<=100000 bonus=i*0.1. /利润在10万元以 内按10%提成奖金 else if(i<=200000 bonus=bon1+(100000·0.075:利润在10万元至 20万时的奖金 else if (i<=400000 bonus-bon2+(i20000090.05:6利润在20万元至 40万时的奖金 else if(i<=600000 bonus-bon4+(i-400000°0.03 /利润在40万元至 60万时的奖金 else if(i<=1000000 bonus=bon6+(6000090.015;利润在60万元至 100万时的奖金 else bonus=bon10+(110000010.01;/润在100万元 以上时的奖金 cout<<"bonus="<<bonus<<endl return 0 313题另一解 #include <iostream> using namespace std; int main o Hong i: float bonus, bon 1 bon 2, bon 4 , bon6, bon 10: int c: 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<< enter i: Cn≥>1 C=/100000 if(c>10)c=10; switch(c) Icase 0: bonus=i". 1; break; case 1: bonus=bon 1+(i-100000)0.075; break; case 2- case 3: bonus=bon2+(i-200000 *0.05; break; cdse斗: case 5: bonus=bon4+(i-400000*0.03: break case 6: case 7. case 8 case 9: bonus=bon6+(i-600000) *0.015; break; case10: bonus=bon10+(-1000000)*0.01 cout<< bonus="<<bonus<<endl return 0 314题 #include <iostream> using namespace std int main( int t, a, b, cout<< enter four numbers: cin>>a>>b>>c>>d cout<<"a="<<a<<",b="<b<<" C="<<c<<"d="<<d<<end if (a>b) It=a, a=b; b=t: t=a; a=c; c=t; if (a>d) It=a; a=d; d=t if(b>c) It=b; b=c, c=t: if (b>d) [t=b; b=d; d=t: I