C第三章课后习题答案谭浩强 .doc

3.2题#include <iostream>#include <iomanip>using namespace std;int main ( )float h,r,l,s,sq,vq,vz; const float pi=3.; cout<<"please enter r,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<<setiosflags(ios:fixed)<<setiosflags(ios:right) <<setprecision(2); cout<<"l= "<<setw(10)<<l<<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; 3.3题#include <iostream>using namespace std;int main ()float c,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:"<<c<<endl; return 0;3.4题#include <iostream>using namespace std;int main ( )char c1,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<<endl; cout<<"用cout语句输出结果为:" cout<<c1<<c2<<endl; return 0;3.4题另一解#include <iostream>using namespace std;int main ( )char c1,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<<endl; cout<<"用cout语句输出结果为:" cout<<c1<<","<<c2<<endl; return 0;3.5题#include <iostream>using namespace std;int main ( )char c1,c2; int i1,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:"<<c1<<" , "<<c2<<endl; return 0;3.8题#include <iostream>using namespace std;int main ( ) int a=3,b=4,c=5,x,y; cout<<(a+b>c && b=c)<<endl; cout<<(a|b+c && b-c)<<endl; cout<<(!(a>b) && !c|1)<<endl; cout<<(!(x=a) && (y=b) && 0)<<endl; cout<<(!(a+b)+c-1 && b+c/2)<<endl; return 0; 3.9题include <iostream>using namespace std;int main ( ) int a,b,c; cout<<"please enter three integer numbers:" cin>>a>>b>>c; if(a<b) if(b<c) cout<<"max="<<c; else cout<<"max="<<b; else if (a<c) cout<<"max="<<c; else cout<<"max="<<a; cout<<endl;return 0; 3.9题另一解#include <iostream>using namespace std;int main ( ) int 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max */ cout<<"max="<<max<<endl; return 0; 3.10题#include <iostream>using namespace std;int main ( ) int x,y; cout<<"enter x:" cin>>x; if (x<1) y=x; cout<<"x="<<x<<", y=x="<<y; else if (x<10) / 1x10 y=2*x-1; cout<<"x="<<x<<", y=2*x-1="<<y; else / x10 y=3*x-11; cout<<"x="<<x<<", y=3*x-11="<<y; cout<<endl;return 0;3.11题#include <iostream>using namespace std;int main () float score; char grade; cout<<"please enter score of student:" cin>>score; while (score>100|score<0) cout<<"data error,enter data again." 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 ()long 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(099999):" 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<<endl; /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<<"original order:" switch(place) case 5:cout<<ten_thousand<<","<<thousand<<","<<hundred<<","<<ten<<","<<indiv<<endl; cout<<"reverse order:" cout<<indiv<<ten<<hundred<<thousand<<ten_thousand<<endl; break; 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; break; 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; return 0; 3.13题#include <iostream>using namespace std;int main () long i; /i为利润 float bonus,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<<"enter 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+(i-100000)*0.075; /利润在10万元至20万时的奖金 else if (i<=400000) bonus=bon2+(i-200000)*0.05; /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<=) bonus=bon6+(i-600000)*0.015; /利润在60万元至100万时的奖金 else bonus=bon10+(i-)*0.01; /利润在100万元以上时的奖金 cout<<"bonus="<<bonus<<endl; return 0; 3.13题另一解#include <iostream>using namespace std;int main ()long i; float bonus,bon1,bon2,bon4,bon6,bon10; 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:" cin>>i; c=i/100000; if (c>10) c=10; switch(c) case 0: bonus=i*0.1; break; case 1: bonus=bon1+(i-100000)*0.075; break; case 2: case 3: bonus=bon2+(i-200000)*0.05;break; case 4: case 5: bonus=bon4+(i-400000)*0.03;break; case 6: case 7: case 8: case 9: bonus=bon6+(i-600000)*0.015; break; case 10: bonus=bon10+(i-)*0.01; cout<<"bonus="<<bonus<<endl; return 0;3.14题#include <iostream>using namespace std;int main ()int t,a,b,c,d; cout<<"enter four numbers:" cin>>a>>b>>c>>d; cout<<"a="<<a<<", b="<<b<<", c="<<c<<",d="<<d<<endl; 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<<"the sorted sequence:"<<endl; cout<<a<<", "<<b<<", "<<c<<", "<<d<<endl; return 0; 3.15题#include <iostream>using namespace std;int main ()int p,r,n,m,temp; cout<<"please enter two positive integer numbers n,m:" cin>>n>>m; if (n<m) 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="<<n<<endl; cout<<"LCD="<<p/n<<endl; / p是原来两个整数的乘积 return 0; 3.16题#include <iostream>using namespace std;int main ()char c; int letters=0,space=0,digit=0,other=0; cout<<"enter one line:"<<endl; 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 other+; cout<<"letter:"<<letters<<", space:"<<space<<", digit:"<<digit<<", other:"<<other<<endl; return 0; 3.17题#include <iostream>using namespace std;int main ()int a,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+.="<<sn<<endl; return 0; 3.18题#include <iostream>using namespace std;int main ()float s=0,t=1; int n; for (n=1;n<=20;n+) t=t*n; / 求n! s=s+t; / 将各项累加 cout<<"1!+2!+.+20!="<<s<<endl; return 0; 3.19题#include <iostream>using namespace std;int main ()int i,j,k,n; cout<<"narcissus numbers are:"<<endl; 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<<n<<" " cout<<endl;return 0; 3.20题#include <iostream>using namespace std; int main() const int m=1000; / 定义寻找范围 int k1,k2,k3,k4,k5,k6,k7,k8,k9,k10; int i,a,n,s; for (a=2;a<=m;a+) / a是21000之间的整数，检查它是否为完数 n=0; / n用来累计a的因子的个数 s=a; / s用来存放尚未求出的因子之和，开始时等于a for (i=1;i<a;i+) / 检查i是否为a的因子 if (a%i=0) / 如果i是a的因子 n+; / n加1，表示新找到一个因子 s=s-i; / s减去已找到的因子，s的新值是尚未求出的因子之和 switch(n) / 将找到的因子赋给k1,.,k10 case 1: k1=i; break; / 找出的笫1个因子赋给k1 case 2: k2=i; break; / 找出的笫2个因子赋给k2 case 3: k3=i; break; / 找出的笫3个因子赋给k3 case 4: k4=i; break; / 找出的笫4个因子赋给k4 case 5: k5=i; break; / 找出的笫5个因子赋给k5 case 6: k6=i; break; / 找出的笫6个因子赋给k6 case 7: k7=i; break; / 找出的笫7个因子赋给k7 case 8: k8=i; break; / 找出的笫8个因子赋给k8 case 9: k9=i; break; / 找出的笫9个因子赋给k9 case 10: k10=i; break; / 找出的笫10个因子赋给k10 if (s=0) / s=0表示全部因子都已找到了 cout<<a<<" is a 完数"<<endl; cout<<"its factors are:" if (n>1) cout<<k1<<","<<k2; / n>1表示a至少有2个因子 if (n>2) cout<<","<<k3; / n>2表示至少有3个因子，故应再输出一个因子 if (n>3) cout<<","<<k4; / n>3表示至少有4个因子，故应再输出一个因子 if (n>4) cout<<","<<k5; / 以下类似 if (n>5) cout<<","<<k6; if (n>6) cout<<","<<k7; if (n>7)cout<<","<<k8; if (n>8)cout<<","<<k9; if (n>9)cout<<","<<k10; cout<<endl<<endl; return 0; using namespace std; int main()3.20题另一解#include <iostream> int m,s,i; for (m=2;m<1000;m+) s=0; for (i=1;i<m;i+) if (m%i)=0) s=s+i; if(s=m) cout<<m<<" is a完数"<<endl; cout<<"its factors are:" for (i=1;i<m;i+) if (m%i=0) cout<<i<<" " cout<<endl; return 0; 3.20题另一解#include <iostream>using namespace std;int main() int k11; int i,a,n,s; for (a=2;a<=1000;a+) n=0; s=a; for (i=1;i<a;i+) if (a%i)=0) n+; s=s-i; kn=i; / 将找到的因子赋给k1k10 if (s=0) cout<<a<<" is a 完数"<<endl; cout<<"its factors are:" for (i=1;i<n;i+) cout<<ki<<" " cout<<kn<<endl; return 0; 3.21题#include <iostream>using namespace std;int main() int i,t,n=20; double a=2,b=1,s=0; for (i=1;i<=n;i+) s=s+a/b; t=a; a=a+b; / 将前一项分子与分母之和作为下一项的分子 b=t; / 将前一项的分子作为下一项的分母 cout<<"sum="<<s<<endl; return 0; 3.22题#include <iostream>using namespace std;int main() int day,x1,x2; day=9; x2=1; while(day>0) x1=(x2+1)*2; / 第1天的桃子数是第2天桃子数加1后的2倍 x2=x1; day-; cout<<"total="<<x1<<endl; return 0; 3.23题#include <iostream>#include <cmath>using namespace std;int main() float a,x0,x1; cout<<"enter a positive number:" cin>>a; / 输入a的值 x0=a/2; x1=(x0+a/x0)/2; do x0=x1; x1=(x0+a/x0)/2; while(fabs(x0-x1)>=1e-5); cout<<"The square root of "<<a<<" is "<<x1<<endl; return 0; 3.24题#include <iostream>using namespace std;int main() int i,k; for (i=0;i<=3;i+) / 输出上面4行*号 for (k=0;k<=2*i;k+) cout<<"*" / 输出*号 cout<<endl; /输出完一行*号后换行 for (i=0;i<=2;i+) / 输出下面3行*号 for (k=0;k<=4-2*i;k+) cout<<"*" / 输出*号 cout<<endl; / 输出完一行*号后换行 return 0; 3.25题#include <iostream>using namespace std;int main() char i,j,k; /* i是a的对手;j是b的对手;k是c的对手*/ for (i='X'i<='Z'i+) for (j='X'j<='Z'j+) if (i!=j) for (k='X'k<='Z'k+) if (i!=k && j!=k) if (i!='X' && k!='X' && k!='Z') cout<<"A-"<<i<<" B-"<<j<<" C-"<<k<<endl; return 0; 4.1题#include <iostream>using namespace std;int main() int hcf(int,int); int lcd(int,int,int); int u,v,h,l; cin>>u>>v; h=hcf(u,v); cout<<"H.C.F="<<h<<endl; l=lcd(u,v,h); cout<<"L.C.D="<<l<<endl; return 0; int hcf(int u,int v) int t,r; if (v>u) t=u;u=v;v=t; while (r=u%v)!=0) u=v; v=r; return(v); int lcd(int u,int v,int h) return(u*v/h); 4.2题#include <iostream>#include <math.h>using namespace std;float x1,x2,disc,p,q;int main()void greater_than_zero(float,float); void equal_to_zero(float,float); void smaller_than_zero(float,float); float a,b,c; cout<<"input a,b,c:" cin>>a>>b>>c; disc=b*b-4*a*c; cout<<"root:"<<endl; if (disc>0) greater_than_zero(a,b); cout<<"x1="<<x1<<",x2="<<x2<<endl; else if (disc=0) equal_to_zero(a,b); cout<<"x1="<<x1<<",x2="<<x2<<endl; else smaller_than_zero(a,b); cout<<"x1="<<p<<"+"<<q<<"i"<<endl; cout<<"x2="<<p<<"-"<<q<<"i"<<endl; return 0;void greater_than_zero(float a,float b) /* 定义一个函数，用来求disc>0时方程的根 */ x1=(-b+sqrt(disc)/(2*a); x2=(-b-sqrt(disc)/(2*a); void equal_to_zero(float a,float b) /* 定义一个函数，用来求disc=0时方程的根 */ x1=x2=(-b)/(2*a); void smaller_than_zero(float a,float b) /* 定义一个函数，用来求disc<0时方程的根 */ p=-b/(2*a); q=sqrt(-disc)/(2*a); 4.3题#include <iostream>using namespace std;int main() int prime(int); /* 函数原型声明 */ int n; cout<<"input an integer:" cin>>n; if (prime(n) cout<<n<<" is a prime."<<endl; else cout<<n<<" is not a prime."<<endl; return 0; int prime(int n) int fla