C/C++例程
1.前n项求和？
#include
int main()
{
int n,i,sum;
sum=0;
while((scanf("%d",&n)!=EOF))
{
sum=0;
for(i=0;i<=n;i++)
sum+=i;
printf("%d\n\n",sum);
}
}
2．输入一个小于或等于九位数的数，输出它是几位数，然后单独顺序输出它的位，最后逆序输出它的位
#include
int digtal(int n)
{
return (n > 0)? 1 + digtal(n/10) : 0;
}
int single(int n)
{
return (n > 0)? single(n/10),printf("%d ",n%10):0;
}
int niverted(int n)
{
return (n > 0)? (printf("%d",n%10),niverted(n/10)):0;
}
int main()
{
int n;
while((scanf("%d",&n)!=EOF))
{
printf("%d has %d digtals\n",n,digtal(n));
printf("single out is:");
single(n);
printf("\nInverted order output IS:");
niverted(n);
printf("\n\n");
}
}
3．给定一个半径求其周长及面积
#include
#include
#include
using namespace std;
#define PI 3.1416
float girth(float r)
{
float l;
if (r > 0)
l = 2 * PI * r;
return l;
}
float area(float r)
{
float s;
if (r > 0)
s = PI * r * r;
return s;
}
int main()
{
float r;
cout << "please enter raduis:";
while(cin >> r)
{
cout << "girth:l=" << girth(r) << endl;
cout << "area:s=" << area(r) << endl << '\n';
cout << "please enter raduis:";
}
return 0;
}
4.打印九九乘法表
#include
#include
#include
using namespace std;
void main()
{
int i,j;
for (i = 1; i <= 9; i++)
for(j = 1; j <= i; j++)
{
cout << j << '*' << i << '=' << i * j << '\t';
if (j == i)cout << '\n';
}
}
4．1、程序启动后，在控制台中显示“请输入一个数字（10-20）：”的提示信息，并等待用户输入内容2、读取用户输入的内容，根据用户输入的内容做以下处理：
A、如果用户输入的数字不在 10到 20之间，显示提示“输入不正确，请重新输入数字”
B、如果用户输入的数字正确，依次输出 1到用户输入的数字范围内的数字（使用循环语句输出数字）。如：用户输入 18，在控制台上依次输出 1,2,3..18
3、 C、用户输入“0”，退出程序
#include
#include
using namespace std;
void main()
{
int digit;
cout << "请输入一个数字(10-20)：";
while(cin >> digit)
{
if (digit == 0)
break;
else if ((10 <= digit )&&(digit <= 20))
{
for (int i =1; i <= digit; i++)
{
cout << i;
if (i != digit)
cout << ',';
}
cout << endl;
}
else
cout << "输入不正确，请重新输入数字" << '\n';
cout << '\n';
}
}
5.求任意一点到一条直线的距离
#include
#include
#include
using namespace std;
class line;
class point
{
public:
int x,y;
public:
point(int x1,int y1)
{
x=x1;
y=y1;
};
friend double dist(line l,point p);
};
class line
{
public:
int a,b,c;
public:
line(int a1,int b1,int c1)
{
a=a1;b=b1;c=c1;
};
friend double dist(line l,point p);
};
double dist(line l,point p)
{
double d;
d = abs(l.a*p.x + l.b*p.y + l.c)/(sqrt(l.a*l.a + l.b*l.b));
return d;
}
void main()
{
int x,y;
int a,b,c;
cout << "请输入一个点:";
while(cin >> x >> y)
{
cout << "点:" << "(" << x << "," << y << ")" << endl;
cout << "请输入直线的三个参数:";
cin >> a >> b >> c;
printf("%d,%d,%d\n",a,b,c);
printf("直线：%d*x+%d*y+%d=0\n",a,b,c);//cout << "直线：" << "(" << a << "," << b << "," << c << ")" << endl;
point p1(x,y);
line l1(a,b,c);
cout << "他们间的距离为：distance=" << dist(l1,p1) << endl << endl;
cout << "请输入一个点:";
}
system("pause");
}
6.给定两三角形边长，求其面积和
#include
#include
#include
using namespace std;
class trig
{
double x,y,z;
double area()
{
double d = (x+y+z)/2;
return sqrt(d*(d-x)*(d-y)*(d-z));
}
public:
trig(int i,int j,int k)
{
x = i;
y = j;
z = k;
}
int judge()
{
if ((z < x+y)&&(x < z+y)&&(y < z+x))
return 1;
else
return 0;
};
friend double twoarea(trig a1,trig a2)
{
return (a1.area() + a2.area());
}
};
void main()
{
trig t1(1,4,5),t2(3,4,5);
if (t1.judge()&&t2.judge())
cout << "两三角形的面积和为:area=" << twoarea(t1,t2) << endl;
else
cout << "不能构成三角形" << endl;
}
7.计算储户三家银行的存款数
#include
#include
using namespace std;
class CBank;
class BBank;
class GBank;
class CBank
{
int balance;
public:
CBank(){balance = 0;}
CBank(int i){balance = i;}
void getblance()
{
cout << "输入中国银行存款数：";
cin >> balance;
}
void disp()
{
cout << "中国银行存款数为：" << balance << endl;
}
friend void total(CBank,BBank,GBank);
};
class BBank
{
int balance;
public:
BBank(){balance = 0;}
BBank(int i){balance = i;}
void getblance()
{
cout << "输入工商银行存款数：";
cin >> balance;
}
void disp()
{
cout << "工商银行存款数为：" << balance << endl;
}
friend void total(CBank,BBank,GBank);
};
class GBank
{
int balance;
public:
GBank(){balance = 0;}
GBank(int i){balance = i;}
void getblance()
{
cout << "输入农业银行存款数：";
cin >> balance;
}
void disp()
{
cout << "农业银行存款数为：" << balance << endl;
}
friend void total(CBank,BBank,GBank);
};
void total(CBank A,BBank B,GBank C)
{
cout << "总款数：" << A.balance + B.balance + C.balance << endl;
}
void main()
{
CBank A;
BBank B;
GBank C;
A.disp();
B.disp();
C.disp();
while(1)
{
A.getblance();
B.getblance();
C.getblance();
total(A,B,C);
cout << endl;
}
}
8.经典测试题，强有力的说明了定义类时、构造函数、析构函数等发生的操作，相当经典，有相当的震撼力
#include
#include
#include
#include
using namespace std;
class B
{
int x,y;
public:
B() { x=y=0;cout << "Construcor1" << endl; }
B(int i) { x=i;y=0;cout << "Construcor2" << endl;}
B(int i,int j) { x=i; y=j; cout << "Construcor3" << endl;}
~B() { cout << "Construcor" << endl; }
void print()
{
cout << "x=" << x << ",y=" << y << endl;
}
};
void main()
{
B *ptr;
ptr=new B[3];
ptr[1]=B(5);
ptr[2]=B(2,3);
for (int i = 0; i < 3; i++)
ptr[i].print();
delete[] ptr;
}
执行结果：
9.简单时钟设计
#include
#include
#include
using namespace std;
class CDate
{
int m_nDay;
int m_nMonth;
int m_nYear;//定义年月日的数据成员
int sec,min,hour;//定义时间数据成员
public:
CDate();
CDate(int day,int month,int year);
CTime(int s,int m,int h);
void Display();
void AddDay();
void AddTime();
void SetDate(int day,int month,int year);
void SetTime(int s,int m,int h);
~CDate();
private:
bool IsLeapYear();
};
CDate::CDate(){}
CDate::CDate(int year,int month,int day)
{
m_nDay = day;
m_nMonth = month;
m_nYear = year;
}
CDate::CTime(int s,int m,int h)
{
sec = s;
min = m;
hour = h;
}
void CDate::Display()
{
char day[3];
char month[3];
char year[5];
char secd[3],mind[3],hourd[3];
_itoa(m_nDay,day,10);
_itoa(m_nMonth,month,10);
_itoa(m_nYear,year,10);
_itoa(sec,secd,10);
_itoa(min,mind,10);
_itoa(hour,hourd,10);
printf("当前日期是：%s/%s/%s\n",day,month,year);
if ((sec < 10)&&(min < 10)&&(hour < 10))//为了时间前面加个0，我容易吗我？
printf("当前时间是：0%s:0%s:0%s",hourd,mind,secd);
else if ((sec < 10)&&(min < 10)&&(hour >= 10))
printf("当前时间是：%s:0%s:0%s",hourd,mind,secd);
else if ((sec < 10)&&(min >= 10)&&(hour < 10))
printf("当前时间是：0%s:%s:0%s",hourd,mind,secd);
else if ((sec >= 10)&&(min < 10)&&(hour < 10))
printf("当前时间是：0%s:0%s:%s",hourd,mind,secd);
else if ((sec < 10)&&(min >= 10)&&(hour >= 10))
printf("当前时间是：%s:%s:0%s",hourd,mind,secd);
else if ((sec >= 10)&&(min < 10)&&(hour >= 10))
printf("当前时间是：%s:0%s:%s",hourd,mind,secd);
else if ((sec >= 10)&&(min >= 10)&&(hour < 10))
printf("当前时间是：0%s:%s:%s",hourd,mind,secd);
else
printf("当前时间是：%s:%s:%s",hourd,mind,secd);
}
void CDate::AddTime()
{
sec++;
if (sec > 59)
{
sec = 00;
min++;
if (min > 59)
{
min = 00;
hour++;
if (hour > 23)
{
hour = 00;
AddDay();
}
}
}
return;
}
void CDate::AddDay()
{
m_nDay++;
if (IsLeapYear())
{
if ((m_nMonth == 2)&&(m_nDay == 30))
{
m_nMonth++;
m_nDay = 1;
return;
}
}
else
{
if ((m_nMonth == 2)&&(m_nDay == 29))
{
m_nMonth++;
m_nDay = 1;
return;
}
}
if (m_nDay > 31)
{
if (m_nMonth == 12)
{
m_nYear++;
m_nMonth = 1;
m_nDay = 1;
}
else
{
m_nMonth++;
m_nDay = 1;
}
}
}
void CDate::SetDate(int year,int month,int day)
{
m_nDay = day;
m_nMonth = month;
m_nYear = year;
}
void CDate::SetTime(int h,int m,int s)
{
sec = s;
min = m;
hour = h;
}
CDate::~CDate(){ }
bool CDate::IsLeapYear()
{
bool bLeap;
if ((m_nYear % 4 == 0)&&(m_nYear % 100 != 0)||(m_nYear % 400 == 0))
bLeap = 1;
else
bLeap = 0;
return bLeap;
}
void main()
{
CDate d;
d.SetDate(2012,10,20);//设置日期
d.SetTime(15,05,00);//设置时间
while(1)
{
d.Display();//显示
d.AddTime();//时间计数
Sleep(1000);//延时一秒钟
system("cls");//刷屏
}
}
10．求两个复数的算术运算（利用运算符重载）
#include
//#include
#include
//using namespace std;
class complex
{
float real,imag;
public:
complex(float r,float i) {real = r;imag = i;}
complex() {real = 0;imag = 0;}
void print();
friend complex operator+(complex a,complex b);
friend complex operator-(complex a,complex b);
friend complex operator*(complex a,complex b);
friend complex operator/(complex a,complex b);
};
void complex::print()
{
cout << real;
if (imag > 0) cout << "+";
if (imag != 0) cout << imag << "i" << endl;
}
complex operator+(complex a,complex b)
{
complex temp;
temp.real = a.real + b.real;
temp.imag = a.imag + b.imag;
return temp;
}
complex operator-(complex a,complex b)
{
complex temp;
temp.real = a.real - b.real;
temp.imag = a.imag - b.imag;
return temp;
}
complex operator*(complex a,complex b)
{
complex temp;
temp.real = a.real * b.real - b.imag*b.imag;
temp.imag = a.real * b.imag - a.imag*b.real;
return temp;
}
complex operator/(complex a,complex b)
{
complex temp;
float tt;
tt = 1/(b.real*b.real + b.imag*b.imag);
temp.real = (a.real*b.real + a.imag*b.imag)*tt;
temp.imag = (b.real*a.imag - a.real*b.imag)*tt;
return temp;
}
void main()
{
complex c1(2.3,4.6),c2(3.6,2.8),c3;
c1.print();
c2.print();
c3 = c1 + c2;
c3.print();
c3 = c1 - c2;
c3.print();
c3 = c1 * c2;
c3.print();
c3 = c1 / c2;
c3.print();
}
11．编写一个学生和教师的数据输入和显示程序，学生数据有编号、姓名、班级和成绩，教师数据有编号、姓名、职称和部门。要求将编号、姓名输入和显示设计成一个类person，并作为学生数据操作类student和教师数据操作类teacher的基类
#include
#include
#include
using namespace std;
class person
{
int number;
char name[10];
public:
void input()
{
cout << "输入编号：";cin >> number;
cout << "输入姓名：";cin >> name;
}
void disp()
{
cout << " 编号" << number << endl;
cout << " 姓名" << name << endl;
}
};
class student:public person
{
char clas[6];
int score;
public:
void input()
{
person::input();
cout << " 班号:"; cin >> clas;
cout << " 成绩:"; cin >> score;
}
void disp()
{
person::disp();
cout << " 班号:" << clas << endl;
cout << " 成绩:" << score << endl;
}
};
class teacher:public person
{
char title[10];
char branch[10];
public:
void input()
{
person::input();
cout << " 职称:"; cin >> title;
cout << " 部门:"; cin >> branch;
}
void disp()
{
person::disp();
cout << " 职称:" << title << endl;
cout << " 部门:" << branch << endl;
}
};
void main()
{
student s1;
teacher t1;
cout << "输入一个学生数据：\n";s1.input();
cout << "输入一个教师数据：\n";t1.input();
cout << "显示一个学生数据：\n";s1.disp();
cout << "显示一个教师数据：\n";t1.disp();
system("pause");
}