【项目4-点和距离】
读程序,写出函数的定义,注意其中枚举类型的用法
enum SymmetricStyle {axisx,axisy,point};//分别表示按x轴, y轴, 原点对称
struct Point{
double x; // 横坐标
double y; // 纵坐标
};
double distance1(Point p1, Point p2); // 两点之间的距离,如果用distance,将会与命名空间std中也已经定义的distance函数重名
double distance0(Point p1);
Point symmetricAxis(Point p,SymmetricStyle style); //返回对称点
int main( ){
Point p1={1,5},p2={4,1},p;
cout<<"两点的距离为:"<<distance1(p1,p2)<<endl;
cout<<"p1到原点的距离为:"<<distance0(p1)<<endl;
p=symmetricAxis(p1,axisx);
cout<<"p1关于x轴的对称点为:"<<"("<<p.x<<", "<<p.y<<")"<<endl;
p=symmetricAxis(p1,axisy);
cout<<"p1关于y轴的对称点为:"<<"("<<p.x<<", "<<p.y<<")"<<endl;
p=symmetricAxis(p1,point);
cout<<"p1关于原点的对称点为:"<<"("<<p.x<<", "<<p.y<<")"<<endl;
return 0;
}
// 求两点之间的距离
double distance1(Point p1,Point p2)
{
double d;
……
return d;
}
// 求点到原点的距离
double distance0(Point p)
{
double d;
……
return d;
}
// 求对称点
Point symmetricAxis(Point p1,SymmetricStyle style)
{
Point p;
……
return p;
}
参考解答:
using namespace std;
enum SymmetricStyle {axisx,axisy,point};//分别表示按x轴, y轴, 原点对称
struct Point
{
double x; // 横坐标
double y; // 纵坐标
};
double distance1(Point p1, Point p2); // 两点之间的距离
double distance0(Point p1);
Point symmetricAxis(Point p,SymmetricStyle style); //返回对称点
int main( )
{
Point p1={1,5},p2={4,1},p;
cout<<"两点的距离为:"<<distance1(p1,p2)<<endl;
cout<<"p1到原点的距离为:"<<distance0(p1)<<endl;
p=symmetricAxis(p1,axisx);
cout<<"p1关于x轴的对称点为:"<<"("<<p.x<<", "<<p.y<<")"<<endl;
p=symmetricAxis(p1,axisy);
cout<<"p1关于y轴的对称点为:"<<"("<<p.x<<", "<<p.y<<")"<<endl;
p=symmetricAxis(p1,point);
cout<<"p1关于原点的对称点为:"<<"("<<p.x<<", "<<p.y<<")"<<endl;
return 0;
}
// 求两点之间的距离
double distance1(Point p1,Point p2)
{
double d;
d=sqrt((p1.x-p2.x)*(p1.x-p2.x)+(p1.y-p2.y)*(p1.y-p2.y));
return d;
}
// 求点到原点的距离
double distance0(Point p)
{
double d;
d=sqrt(p.x*p.x+p.y*p.y);
return d;
}
// 求对称点
Point symmetricAxis(Point p1,SymmetricStyle style)
{
Point p;
p.x=p1.x;
p.y=p1.y;
switch(style)
{
case axisx:
p.y=-p1.y; break;
case axisy:
p.x=-p1.x; break;
case point:
p.x=-p1.x;p.y=-p1.y;
}
return p;
}
感谢单昕昕发现distance在命名上的问题。