网易云音乐链接 : 点我听歌.
1、定义复数类的加法与减法，使之能够执行下列运算：
complex a(2,5),b(7,8),c(0,0);
c=a+b;
c=4.1+a;
c=b-5.6;
代码
#include <iostream.h>
#include <string.h>
//using namespace std;
class Complex
{
double real;
double imag;
public:
Complex()
{real=0.0; imag=0.0;}
Complex(double r, double i)
{real = r; imag = i;}
void show()
{cout<<real<<" + "<<imag<<"i"<<endl;}
Complex operator+(const Complex &y) //做为成员函数-对象加法
{
Complex t(0,0);
t.real = real + y.real;
t.imag = imag + y.imag;
return t;
}
Complex operator-(const Complex &y1) //做为成员函数-对象减法
{
Complex t(0,0);
t.real = real - y1.real;
t.imag = imag - y1.imag;
return t;
}
Complex operator+(const double v)
{
Complex t(0,0);
t.real = v + real;
t.imag = imag;
return t;
}
Complex operator-(const double v1)
{
Complex t(0,0);
t.real = v1 - real;
t.imag = -imag;
return t;
}
friend Complex operator-(double, Complex &);
friend Complex operator+(double, Complex &);
};
Complex operator+(double c1, Complex &c2)
{
Complex c;
c.real=c2.real+c1;
c.imag=c2.imag;
return c;
}
Complex operator-(double c1, Complex &c2)
{
Complex c;
c.real=c1-c2.real;
c.imag=-c2.imag;
return c;
}
void main()
{
Complex a(2, 5);Complex b(7, 8);Complex c;
c = a + b;
c.show();
c = 4.1 + a;
c.show();
c=b-5.6;
c.show();
}
运行结果
◇ 2、设计一个2行3列的矩阵类，重载运算符“+”和“－”，能实现矩阵类对象的加减运算；重载流插入运算符“<<”和流提取运算符“>>”,使之能用于矩阵的输入和输出。
代码
#include <iostream.h>
class matrix
{
private:
int a[2][3];
public:
friend matrix operator+(matrix&A, matrix&B);
friend ostream& operator<<(ostream&, matrix&);
friend istream& operator>>(istream&, matrix&);
matrix(){};
};
matrix operator+(matrix&A,matrix&B)
{
matrix c;
for(int i=0;i<2;i++)
for(int j=0;j<3;j++)
c.a[i][j]=A.a[i][j]+B.a[i][j];
return c;
}
istream & operator >> (istream &input,matrix &a)
{
cout<<"请输入矩阵： "<<endl;
for(int i=0;i<2;i++)
for(int j=0;j<3;j++)
input>>a.a[i][j];
return input;
}
ostream & operator << (ostream &output,matrix &C)
{
cout<<"矩阵相加的结果为："<<endl;
for(int i=0;i<2;i++)
{
for(int j=0;j<3;j++)
output<<C.a[i][j]<<" ";
cout<<endl;
}
return output;
}
void main()
{
matrix A,B,C;
cin>>A;
cin>>B;
C=A+B;
cout<<C;
}
运行结果
◇ 3、定义一个人民币的类，其中成员数据包括元、角、分，成员函数包括构造及输出函数。要求增加适当的成员函数，重载“+”“－”“+＝”“++”及输入输出流，实现人民币的直接运算。注意分、角、元的进位。
代码
#include <iostream.h>
#include <string.h>
#include <iomanip.h>
class RMB
{
int yuan, jiao ,fen;
public:
RMB(int y=0, int j=0, int f=0)
{
yuan = y; jiao = j; fen = f;
}
void show()
{
cout<<yuan<<"."<<jiao<<fen<<endl;
}
RMB operator+(RMB &r); //成员函数重载加运算符
friend RMB operator-(RMB &r1, RMB &r2); //用友元函数形式重载减
RMB& operator+=(RMB &r); //只能用成员函数形式重载+=
RMB& operator++(); //成员函数形式重载++
friend istream &operator>>(istream &input, RMB &);
friend ostream &operator<<(ostream &output, RMB &); //只能用友元重载输入输出流
};
RMB RMB::operator+(RMB &r)
{
RMB rs;
rs.fen=fen+r.fen;
while(rs.fen>=10)
{
rs.jiao++;
rs.fen-=10;
}
rs.jiao=jiao+r.jiao;
while(rs.jiao>=10)
{
rs.yuan++;
rs.jiao-=10;
}
rs.yuan=yuan+r.yuan;
return rs;
}
RMB operator-(RMB &r1, RMB &r2)
{
RMB rs;
if(r1.yuan<r2.yuan)
{
cout<<"钱不够，不能付账!
";
return rs;
}
if(r1.yuan==r2.yuan&&r1.jiao<r2.jiao)
{
cout<<"钱不够，不能付账!
";
return rs;
}
if(r1.yuan==r2.yuan&&r1.jiao==r2.jiao&&r1.fen<r2.fen)
{
cout<<"钱不够，不能付账!
";
return rs;
}
rs.yuan=r1.yuan-r2.yuan;
rs.jiao=r1.jiao-r2.jiao;
while(rs.jiao<0)
{
rs.yuan--;
rs.jiao+=10;
}
rs.fen=r1.fen-r2.fen;
while(rs.fen<0)
{
rs.jiao--;
if(rs.jiao<0)
{
rs.yuan--;
rs.jiao+=10;
}
rs.fen+=10;
}
return rs;
}
RMB& RMB::operator+=(RMB &r) //返回对象本身的引用
{
fen=fen+r.fen;
while(fen>=10)
{
jiao++;
fen-=10;
}
jiao=jiao+r.jiao;
while(jiao>=10)
{
yuan++;
jiao-=10;
}
yuan=yuan+r.yuan;
return *this;
}
RMB& RMB::operator++(void) //对象本身加
{
fen=fen+1;
while(fen>=10)
{
jiao++;
fen-=10;
}
while(jiao>=10)
{
yuan++;
jiao-=10;
}
return *this;
}
istream &operator>>(istream &input, RMB &r)
{
cout<<"请输入元、角、分（整数之间用空格或回车符间隔）:";
input>>r.yuan>>r.jiao>>r.fen;
return input;
}
ostream &operator<<(ostream &output, RMB &r) //只能用友元重载输入输出流
{
while(r.fen>=10)
{
r.jiao++;
r.fen-=10;
}
while(r.jiao>=10)
{
r.yuan++;
r.jiao-=10;
}
output<<"￥"<<setw(3)<<r.yuan<<"."<<r.jiao<<r.fen;//输出到输出流的引用
return output;
}
void main()
{
RMB r1(14, 5, 7),r2,r3;
cout<<"请输入一个人民币数值：
";
cin>>r2;
r3=r1+r2;
cout<<r1<<" + "<<r2<<" = "<<r3<<endl;
cout<<r3<<" + "<<r1<<" = ";
r3+=r1;
cout<<r3<<endl;
cout<<r3<<" - "<<r1<<" = ";
r3=r3-r1;
cout<<r3<<endl;
cout<<r3<<" + "<<"￥ 0.01=";
++r3;
cout<<r3<<endl;
}
运行结果
◇ 4、定义一个时间的类，其中成员数据包括小时、分、秒，成员函数为构造函数。要求增加适当的成员函数，重载“+”“－”“+＝”及输入输出流，实现时间类对象的直接输入输出及两个时间的运算。
代码
#include <iostream.h>
#include <iomanip.h>
#include <time.h>
class TIME
{
int hour, minute, second;
public:
TIME(int h=0, int m=0, int s=0)
{
hour=h; minute=m; second=s;
}
TIME operator+(TIME &t); //成员函数形式重载
friend double operator-(TIME &t1, TIME &t2);
TIME& operator+=(TIME &t); //只能用成员函数
//TIME& operator++();
friend istream &operator>>(istream &input, TIME &);
friend ostream &operator<<(ostream &output, TIME &);
};
TIME TIME::operator+(TIME &t) //成员函数形式重载
{
TIME ot;
ot.second=second+t.second;
while(ot.second>=60)
{
ot.minute++;
ot.second-=60;
}
ot.minute+=minute+t.minute;
while(ot.minute>=60)
{
ot.hour++;
ot.minute-=60;
}
ot.hour=hour+t.hour;
if(ot.hour>=24)
ot.hour-=24;
return ot;
}
double operator-(TIME &t1, TIME &t2)
{
double minute;
if((t1.hour<t2.hour)||(t1.hour==t2.hour&&t1.minute<t2.minute)||(t1.hour==t2.hour&&t1.minute==t2.minute
&&t1.second<t2.second))
{
cout<<"不能相减!
";
return 0;
}
minute=(t1.hour-t2.hour)*60;
minute=t1.minute-t2.minute;
if(t1.second-t2.second)
minute--;
return minute;
}
TIME& TIME::operator+=(TIME &t) //只能用成员函数
{
second=second+t.second;
if(second>60)
{
minute++;
second-=60;
}
minute=minute+t.minute;
while(minute>=60)
{
hour++;
minute-=60;
}
hour=hour+t.hour;
if(hour>=24)
hour=hour-24;
return *this;
}
//TIME& operator++();
istream &operator>>(istream &input, TIME &t)
{
cout<<"请输入时间（按tip输入小时、分、秒）:
";
cout<<"请输小时（大于0，小于等于23）:";
input>>t.hour;
while(t.hour>23||t.hour<0)
{
cout<<"输入的小时没按要求输入！请重新时输入！";
input>>t.hour;
}
cout<<"请输分钟（大于等于0，小于等于59）:";
input>>t.minute;
while(t.minute>59||t.minute<0)
{
cout<<"输入的分钟没按要求输入！请重新时输入！";
input>>t.minute;
}
cout<<"请输秒数（大于等于0，小于等于59）:";
input>>t.minute;
while(t.second>59||t.second<0)
{
cout<<"输入的秒数没按要求输入！请重新时输入！";
input>>t.second;
}
return input;
}
ostream &operator<<(ostream &output, TIME &t)
{
output<<setw(2)<<t.hour<<":"<<t.minute<<":"<<t.second;
return output;
}
void main()
{
struct tm *CurrentTime;
time_t time_date(0); //两个变量获取当前时间
CurrentTime=localtime(&time_date);//设置当前时间
TIME currenttime(CurrentTime->tm_hour,CurrentTime->tm_min,CurrentTime->tm_sec);
//将当前时间赋值给时间类对象
cout<<"
欢迎使用简易时间计算器！
";
cout<<"
当前时间是："<<currenttime<<endl;
int Choice; //输入选择用
TIME t1, t2, t3;
double m;
do
{
cout<<"
1.时间加法
";
cout<<"
2.时间减法
";
cout<<"
3.时间自身加法（+=）
";
cout<<"
0.退出时间计算器
";
cout<<"
请选择你需要的操作：";
cin>>Choice;
switch(Choice)
{
case 1:
{
cout<<" 时间加法，请按提示输入两个时间:
";
cin>>t1;
cout<<endl;
cin>>t2;
t3=t1+t2;
cout<<endl<<t1<<" + "<<t2<<" = "<<t3<<endl;
break;
}
case 2:
{
cout<<" 时间减法，请按提示输入两个时间:
";
cin>>t1;
cout<<endl;
cin>>t2;
m=t1-t2;
cout<<endl<<t1<<" - "<<t2<<" = "<<m<<"分钟"<<endl;
break;
}
case 3:
{
cout<<" 时间自身加法，请按提示输入两个时间:
";
cin>>t1;
cout<<endl;
cin>>t2;
t3=t1+t2;
cout<<endl<<t1<<" += "<<t2;
t1+=t2;
cout<<t1<<endl;
break;
}
case 0:
{
break;
}
default://未选择0~3数字，重新选择
cout<<"输入错误，请重新输入你的
";
break;
}
}while(Choice != 0);
cout<<"
欢迎再次使用简易时间计算器，谢谢！"<<endl;
}
运行结果
◇ 5、用弦截法求以下方程的根。
(1) f1(x)=xxx+xx-3x+1,初值为x1=0.5,x2=1.5。
(2) f2(x)=xx-2x-8,初值为x1=-3,x2=3。
(3) f3(x)=xxx+2x*x+2x+1，初值为x1=-2,x2=3。
(1) f1(x)=xxx+xx-3x+1,初值为x1=0.5,x2=1.5*
代码1
#include<iostream>
#include<iomanip>
#include<cmath>
using namespace std;
double f(double); //函数声明
double xpoint(double,double); //函数声明
double root(double,double); //函数声明
int main( )
{
double x1,x2,f1,f2,x;
do
{ cout<< "input x1,x2:";
cin >> x1 >> x2;
f1=f(x1);
f2=f(x2);
} while(f1*f2 >= 0);
x = root(x1,x2);
cout << setiosflags(ios::fixed) << setprecision(7); //指定输出7位小数
cout << "A root of equation is " << x << endl;
return 0;
}
double f(double x) //定义f函数
{ double y;
y = x * x * x + x * x-3*x+1;
return y;
}
double xpoint(double x1,double x2) //定义spoint 函数,求出弦与 x 轴的交点
{ double y;
y = ( x1 * f(x2) - x2 * f(x1))/(f(x2) - f(x1)); //在 xpoint 函数中调用 f 函数
return y;
}
double root(double x1,double x2) //定义 root 函数，求近似根
{
double x,y,y1;
y1 = f(x1);
do
{ x = xpoint(x1,x2); //在 root 函数中调用 xpoint 函数
y = f(x); //在 root 函数中调用 f 函数
if(y*y1 > 0)
{ y1 = y;
x1 = x;
}
else
x2 = x;
}while(fabs(y)>= 0.00001);
return x;
}
运行结果1
(2) f2(x)=x*x-2x-8,初值为x1=-3,x2=3。
代码2
#include<iostream>
#include<iomanip>
#include<cmath>
using namespace std;
double f(double); //函数声明
double xpoint(double,double); //函数声明
double root(double,double); //函数声明
int main( )
{
double x1,x2,f1,f2,x;
do
{ cout<< "input x1,x2:";
cin >> x1 >> x2;
f1=f(x1);
f2=f(x2);
} while(f1*f2 >= 0);
x = root(x1,x2);
cout << setiosflags(ios::fixed) << setprecision(7); //指定输出7位小数
cout << "A root of equation is " << x << endl;
return 0;
}
double f(double x) //定义f函数
{ double y;
y = x*x-2*x-8;
return y;
}
double xpoint(double x1,double x2) //定义spoint 函数,求出弦与 x 轴的交点
{ double y;
y = ( x1 * f(x2) - x2 * f(x1))/(f(x2) - f(x1)); //在 xpoint 函数中调用 f 函数
return y;
}
double root(double x1,double x2) //定义 root 函数，求近似根
{
double x,y,y1;
y1 = f(x1);
do
{ x = xpoint(x1,x2); //在 root 函数中调用 xpoint 函数
y = f(x); //在 root 函数中调用 f 函数
if(y*y1 > 0)
{ y1 = y;
x1 = x;
}
else
x2 = x;
}while(fabs(y)>= 0.00001);
return x;
}
运行结果2
(3) f3(x)=xxx+2x*x+2x+1，初值为x1=-2,x2=3。
代码3
#include<iostream>
#include<iomanip>
#include<cmath>
using namespace std;
double f(double); //函数声明
double xpoint(double,double); //函数声明
double root(double,double); //函数声明
int main( )
{
double x1,x2,f1,f2,x;
do
{ cout<< "input x1,x2:";
cin >> x1 >> x2;
f1=f(x1);
f2=f(x2);
} while(f1*f2 >= 0);
x = root(x1,x2);
cout << setiosflags(ios::fixed) << setprecision(7); //指定输出7位小数
cout << "A root of equation is " << x << endl;
return 0;
}
double f(double x) //定义f函数
{ double y;
y = x*x*x+2*x*x+2*x+1;
return y;
}
double xpoint(double x1,double x2) //定义spoint 函数,求出弦与 x 轴的交点
{ double y;
y = ( x1 * f(x2) - x2 * f(x1))/(f(x2) - f(x1)); //在 xpoint 函数中调用 f 函数
return y;
}
double root(double x1,double x2) //定义 root 函数，求近似根
{
double x,y,y1;
y1 = f(x1);
do
{ x = xpoint(x1,x2); //在 root 函数中调用 xpoint 函数
y = f(x); //在 root 函数中调用 f 函数
if(y*y1 > 0)
{ y1 = y;
x1 = x;
}
else
x2 = x;
}while(fabs(y)>= 0.00001);
return x;
}
运行结果3
