C程序设计习题

1、C+程序设计 练习题学院：计算机学院专业班级： 物联网1002学号：姓名： 徐远志第八章1.下面是一个类的测试程序,试设计出能是用如下测试程序的类Int main()Test x;(30,20);()；return 0;解：#in clude <iostream>using n amespace std;class Test义并实现一个矩形类Crectangle.该类包含了以下成员函数.Crecta ngle():累的构造函数,根据需要可以定义多个构造函数SetTop(),SetLeft():设置矩形的左上角坐标SetLe ngth(),SetWidth():设置矩形的长和宽Per

2、imeter():求矩形的周长Area():求矩形的面积GetWidth():返回矩形的宽度Getle ngth():返回矩形的长度IsSquare():判断矩形是否为正方形Move():将矩形从一个位置移动到另一个位置Size():改变矩形的大小Where():返回矩形的左上角的坐标PrintRectangle():输出矩形的四个顶点的坐标数据成员int top,left;int length,width;解：#include <iostream>using namespace std;class Crecta ngle次歌手大赛中,有JudgeNum个评委给选手打分,参加比赛的

3、选手有 PlayerNum名,现为比赛积分编写一个 CompetitionResult 类,类的定义如下：(定义略)(1)写出所有的成员函数和实现代码.(2) 编写mai n ()函数对该类进行测试.在函数体中,定义了一个competiti on Result 类的对象数组 rPlauerNum, 他的每个元素记录了每个选手的所有信息, 个评委的 打分通过键盘的输入,在屏幕上应有提示信息进行交互式操作,比赛结果按选手 得分从高到底排列输出.解：( 1) CompetitionResult:CompetitionResult()num=0; strcpy(name,"");f

4、or(int i=0;i<JudgeNum;i+)scorei=;average=0;CompetitionResult:CompetitionResult(short n,char*ps)num=n;strcpy(name,ps);for(int i=0;i<JudgeNum;i+)scorei=;average=0;float CompetitionResult:MaxScore() int max=score0;for(int i=0;i<JudgeNum;i+)if(max<scorei) max=scorei;return max;float Competit

5、ionResult:MinScore() int min=score0;for(int i=0;i<JudgeNum;i+) if(min>scorei) min=scorei; return min;void CompetitionResult:SetAvg() int i;float sum=;for(i=0;i<JudgeNum;i+)sum+=scorei;sum=sum-MaxScore()-MinScore(); average=sum/(JudgeNum-2);float CompetitionResult:GetAvg() return average;sho

6、rt CompetitionResult:GetNo(int i) return num;void CompetitionResult:SetNo(int i) num=i;char CompetitionResult:* GetName() return name ; float CompetitionResult:GetScore(int j) return scorej; void CompetitionResult:Setscore(int k,float av) scorek=av; void Sort(CompetitionResult *pr,int n)int i,j ,k;C

7、ompetitionResult temp; for(i=0;i<n-1;i+)k=i;for(j=i+1;j<n;j+) if(prj.average)>(prk.average) k=j;if(k!=j)temp=pri;pri=prk; prk=temp;(2)int main()CompetitionResult playerPlayerNum=CompetitionResult(1,"one"),CompetitionResult(2,"two"),CompetitionResult(3,"there"),

8、CompetitionResult(4,"four"),CompetitionResult(5,"five"),CompetitionResult(6,"six"),CompetitionResult(7,"seven"),CompetitionResult(8,"eight"),CompetitionResult(9,"nine"),CompetitionResult(10,"ten");float a;for(int i=0;i<PlayerNu

9、m;i+)cout<<" 请评委给第 "<<i+1<<" 个选手打分 :"<<endl;for(int j=0;j<JudgeNum;j+) cin>>a;playeri.Setscore(j,a);playeri.SetAvg();cout<<endl;Sort(player,PlayerNum);cout<<" 最后的得分情况为： "<<endl;for (int j=0;j<PlayerNum;j+)cout<<

10、;j+1<<" "cout<<setw(6)<<playerj.GetAvg();cout<<" "<<playerj.GetName()<<endl;return 0;程序运行的结果为：第九章1.在下面的程序中,派生类Derived的成员函数sets ()能否访问基类base 中得变量 a 和 b 为什么如果在基类 base 中将数据成员 a 和 b 定义为私有成 员,下面的程序能否通过编译为什么#include <iostream>using namespace st

11、d;class Base public:void set(int i,int j) a=i;b=j;void show() cout<<"a="<<a<<",b="<<b<<endl; protected:int a,b;class Derived:public Base public:void sets() s=a*b;void shows() cout<<"s="<<s<<endl;private:int s;int main() De

12、rived obj;(2,3);();();();return 0;解：可以访问a和b;由于a, b为保护成员,经公有继承,成为派生类中得保护成员,所以派生类Derived的成员函数sets ()能访问基类base中得 变量 a 和 b.如果定义为私有,那么不能通过编译.由于派生类中的成员函数不能访问 基类中得私有成员2.声明一个图形基类Shape,在它的根底上派生出矩形类 Rectangle和圆形 类 Circle ,它们都有计算面积的和周长、输出图形信息的成员函数,再在 Rectangle类的根底上派生方形类Square.编写程序和各类的定义和实现, 以及类的使用.解：#include &

13、lt;iostream>using namespace std;class Shape public:double getArea()double getPerimeter() ;class Rectangle:public Shape protected:double height;double width;public:Rectangle() Rectangle(double a,double b)height=a;width=b;double getArea() return height*width;double getPerimeter() return 2*(height+w

14、idth);class Circle:public Shapepublic:Circle(double x):r(x)double getPerimeter()return 2*r;double getArea()return r*r*;private:double r;class Square:public Rectanglepublic:Square(double x)height=x;width=x;int main()Rectangle a1,;Circle a2;Square a3;cout<<"Area="<<()<<endl

15、; cout<<"Perimeter="<<()<<endl; cout<<"Area="<<()<<endl;cout<<"Perimeter="<<()<<endl; cout<<"Area="<<()<<endl;cout<<"Perimeter="<<()<<endl;return 0;5. 某销售公司有销售经

16、理和销售员工,月工资的计算方法为：销售鼓励的底 薪为 4000 元,并将销售额的 2/1000 做提成,销售员工无底薪,只提取 5/1000 为工资.编写程序：(1) 定义一个积累 Employee,它包含三个数据成员number, name和 salary ,以及编号和姓名的构造函数.(2) 有Employee类派生Salesman类.Salesman类包含两个新数据成员 commrate和sales,还包括了用于输入销售额并计算销售员工工资 的成员函数pay ()和用于输出打印的print ().( 3) 由 Salesman 派生 Salemanager,Salesmmanage 类包含

17、新数据成员 monthlypay,以及用于输入销售额并计算销售经理工资的成员函数 pay ()和用于输出的print ().(4)编写main ()函数中测试所设计的类结构.解：(1) class Employeepublic:Employee(long num,char *n)number=num;strcpy(name,n);protected:long number;char name20;double salary;( 2)class Salesman:public Employeepublic:Salesman(long num,char *n,float c=:Employee(n

18、um,n) commrate=c;void pay()coutvv" ee "<<name<<"士 ie 3eue :"cin>>sales;salary=commrate*sales;void print()cout<<name«"a 10xe ao"<<salary<<endl;protected:float commrate;double sales;(3)class Salesmanager:public Salesmanpublic:c=,do

19、ubleSalesmanager(long num,char *n,floatm=4000):Salesman(num,n,c)monthlypay=m;void pay()cout«" ee "<<name<<"士 ie 3eotl; e£cin>>sales;salary=monthlypay+commrate*sales;void print()cout<<name«"a 10xe ao"<<salary<<endl;private:do

20、uble monthlypay;(4) int main()Salesman s1(1234," ae e");Salesmanager s2(1357,"");();();();();return 0;第十章1.定义一个辅助类Complex,重载运算符+、-、*、/使之能用于复数的加减乘 除运算.运算符重载函数为 Complex类的成员函数.编写程序,分别求出两 个复数之和、差、积、商.解：#include <iostream>using namespace std;class Complexpublic:Complex(double r=

21、,double i= real=r;imag=i;Complex operator + (Complex);friend Complex operator - (Complex,Complex);Complex operator * (Complex);friend Complex operator / (Complex,Complex); void display();private:double real,imag;Complex Complex:operator + (Complex c)return Complex(real+,imag+;Complex operator - (Com

22、plex c1,Complex c2) return Complex Complex:operator * (Complex c) return Complex(real*+,imag*+real*;Complex operator / (Complex c1,Complex c2)double r,i;r=*+*/*+*;return(r,i);void Complex:display()cout<<"("<<real<<","<<imag<<")"<<endl

23、;int main()Complex c1,c2,5),c3;cout<<"c1=" ();cout<<"c2=" ();c3=c1+c2;cout<<"c1+c2="();c3=c1-c2;cout<<"c1-c2="();c3=c1*c2;cout<<"c1*c2="();c3=c1/c2;cout<<"c1/c2="();return 0;2.定义一个日期类Date,包括年,月,日等私有数据成员.要

24、求为所定义的Date 类设计如下重载运算符函数：Date operator+(int days); 返回一日起加天数 days 后得到的日期Date operator- ( int days); 返回一日起减天数 days 后得到的日期Int operator- ( Date &b); 返回两日期相差的天数解：#include <iostream>using namespace std;int day_tab12=31,28,31,30,31,30,31,31,30,31,30,31;class Datepublic:Date() Date(int y,int m,int

25、d) year=y;mon=m;day=d;void setday(int d) day=d;void setmon(int m) mon=m;void setyear(int y) year=y;int getday() return day;int getmon() return mon;int getyear() return year;Date operator + (int days)static Date date;int number=dton(*this);number-=days;date=ntod(number);return date;Date operator - (i

26、nt days)static Date date;int number=dton(*this);number-=days;date=ntod(number);return date;int operator - (Date &b)int days=dton(*this)-dton(b)-1;return days;void disp()cout<<year<<"."<<mon<<"."<<day<<endl;private:int year,mon,day;int leap(

27、int);int dton(Date &);Date ntod(int);int Date:leap(int year)if(year%4=0&&year&100!=0|year%400=0)return 1;else return 0;int Date:dton(Date &d)int y,m,days=0;for(y=1;y<=;y+)if(leap(y)days+=366;elsedays+=365;day_tab1=leap29:28;for(m=0;m<m+)days+=day_tabm;days+=;return days;Dat

28、e Date:ntod(int n)int y=1,m=1,d,rest=n,days_year;while(1)days_year=leap(y)366:365;if(rest<days_year)break;elserest-=days_year;y+;y-;day_tab1=leap(y)29:28;while(1)if(rest>day_tabm-1)rest-=day_tabm-1;elsebreak;m+;d=rest;return Date(y,m,d);int main()Date now(2021,12,22),Nationday(1949,10,1);cout&

29、lt;<"now:"();cout<<"Nationday:"();cout«" a 2 lie y:"<<(nNWationday)«endl;Date d1= no w+365,d2二 no w-365;cout<<"now+365:"();cout<<"now-365:"();return 0;3. 编写一个程序尖酸正方体圆柱体的外表积和体积. 要求抽象出一个公共的基类Body,吧他当做抽象类,在该类中定义外表积和体

30、积的纯虚函数.抽象 类中定义一个数据成员 data ,他可以作为求的半径, 正方形的边长, 或者圆 柱体的地面半径.在三个类中都有计算白面积和体积的函数的具体实现.解：#include <iostream>using namespace std;class Bodyprotected:double data;public:Body(double data)Body:data=data;virtual double s()=0;virtual double v()=0;class Cube:public Bodypublic:Cube(double data):Body(data);

31、double s()return data*data*6;double v()return data*data*data;class Sphere:public Bodypublic:Sphere(double r):Body(r);double s()return data*data*4;double v()return 4*data*data*data/3;class Cylinder:public Bodydouble height;public:height):Cylinder(double r,double Body(r)Cylinder:height=height;double s

32、()return (2*data*data+2*data*height);double v()return data*data*height;int main()Body *p;Cube obj1(5);Sphere obj2(5);Cylinder obj3(5,5);p=&obj1;cout<<" 正方体的外表积： "<<p->s()<<endl;cout<<" 正方体的体积： "<<p->v()<<endl;p=&obj2;cout<<&

33、quot; 球的外表积： "<<p->s()<<endl;cout<<" 球的体积： "<<p->v()<<endl;p=&obj3;cout<<" 球的外表积： "<<p->s()<<endl;cout<<" 球的体积： "<<p->v()<<endl; return 0;第十一章1. Void Sort(int a,int size);Void Sort(do

34、uble a,int size);是设计函数模板.是一个函数模板的两个例子, 起功能是将数组 a 中得前 size 个元素按从小到大的顺序依次排列 解： #include <iostream> using namespace std;template<class T> void Sort(T set,int n)int i,j; T temp;for (i=1;i<n;i+) for(j=n-1;j>i;j-)if(setj-1>setj) temp=setj-1;setj-1=setj; setj=temp;int main()int a=4,5,8

35、,9,3;double b=3,5,2,;Sort(a,6);Sort(b,6);for(int i=0;i<6;i+) cout<<ai<<" "cout<<endl; for(i=0;i<6;i+) cout<<bi<<" "cout<<endl; return 0;2. 设有如下声明 class Test public ：Void SetData1(int val) data1=val;Void SetData2(doule val) data2=val;int G

36、etData1() return data1;double GetData2() return data2;private:int data1;double data2;试将此类声明为类模板的声明,使得数据成员 data1 和 data2 可以使任何类型 #include <iostream>#include <string>using namespace std;template <class T1,class T2>class Testpublic:void SetData1(T1 val) data1=val;void SetData2(T2 val)

37、 data2=val;T1 GetData1() return data1;T2 GetData2() return data2;private:int data1;double data2;int main()Test<int,double> t1;(10);Jcout<<()<<" "<<()<<endl;Test<char ,string> t2;('a');("China");cout<<()<<" "<<

38、;()<<endl;return 0;3 栈是一种重要的数据结构,它是一种只允许在彪的一端进行插入或删除操作的线性表.表中允许进行插入、删除操 作的一端称为栈顶.表的令一端称为栈底.下面是一个整形栈类的定义：Const int MaxSize=100;Class IstackPublic:Istack();void Push(int &n);void pop();int GetTop();bool Empty();int size();boid ClearStack();Istack();Private:Int elemMaxSize;Int top;编写一个栈的类模板,以

39、便为任何类型的对象提供栈结构的数据操作. 解：#include <iostream>#include <string>using namespace std;const int MaxSize=100;template <class T>class Stackpublic:Stack();void Push(const T &a);void Pop();T GetTop();void ClearStack();bool Empty();int Size();Stack()private:T elemMaxSize;int Top;template &

40、lt;class T>Stack <T>:Stack()Top=-1;template <class T>void Stack<T>:Push(const T &a)if(Top=MaxSize-1)return;Top+; elemTop=a;template <class T>void Stack<T>:Pop()if(Top=-1) return;Top-;template <class T>void Stack<T>:ClearStack()Top=-1;template <clas

41、s T>T Stack<T>:GetTop()if(Top=-1)return 0; return elemTop;template <class T>bool Stack<T>:Empty()if(Top=-1)return true;elsereturn false;template <class T>int Stack<T>:Size()return Top+1;int mian()Stack<int> s1;Stack<string> s2;(123);(456); cout<<()&

42、lt;<endl;();cout<<()<<" "<<()<<endl;("beijing");("wuhan"); cout<<()<<" "<<()<<endl;return 0;第十三章1. 编写程序,重载运算符 <<和>>,使用户能直接输入和输出复数. 解：#include <>class Complexpublic:Complex(double r=0,double i

43、=0)real=r;imag=i;friend ostream &operator<<(ostream &,Complex &);friend istream &operator>>(istream &,Complex &); private:double real,imag;ostream &operator<<(ostream &output ,Complex &obj)output<<if>0)output<<"+"if!=0)ou

44、tput<<<<"i"return output;istream &operator>>(istream &input ,Complex &obj)cout<<"Input the real and img of the complex:"<<endl; input>>input>>return input;int main()Complex c1,c2,c3; cin>>c1>>c2>>c3;cout<&

45、lt;"c1="<<c1<<endl;cout<<"c2="<<c2<<endl;cout<<"c3="<<c3<<endl;return ;然后3. 编写一个程序, 统计文本文件中的字符, 并给文件所有行加上行号, 写到中去 解：#include <iostream>#include <fstream>using namespace std;int main()fstream file;("",ios:in);if(!file)cout<<" can't open."<<endl;abort();char ch;int i=0;while(!()(ch);i+;cout<<"Charater:"<<i<<endl;();return 0;6#include <iostream>#include <fstream>using namespace std;struct