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C++第2周项目6——点结构体与枚举

2013年06月11日 ⁄ 综合 ⁄ 共 1736字 ⁄ 字号 评论关闭

课程首页地址:http://blog.csdn.net/sxhelijian/article/details/7910565,本周题目链接:http://blog.csdn.net/sxhelijian/article/details/8635385

【项目6-点结构体与枚举】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 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;
}

参考解答:

#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;
}

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