项目5点结构体与枚举

来源：互联网发布：javascript代码规范编辑：程序博客网时间：2024/04/20 08:26

#include "iostream"#include "Cmath"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 distance(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："<<distance(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 distance(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;}

光最后求对称点问题，出了很多次错误。没想到还要再将p1的x,y先付给p.

0 0