m0_59064448 2021-12-05 18:50 采纳率: 55.6%

# 能不能写一个接口？急！

``````#include <iostream>
#include <math.h>
using namespace std;
class Point
{
public:
double x,y;
Point(){x=0;y=0;}
Point(int xx,int yy){x=xx;y=yy;}
};
class Circle
{
public:
Point center;
Circle(){}
Circle(Point p,double r)
{
center = p;
}
void fun(Point p)
{
double dis = sqrt( (center.x-p.x)*(center.x-p.x) + (center.y-p.y)*(center.y-p.y) );
cout << "在圆内"<< endl;
cout << "在圆上"<< endl;
else
cout << "在圆外" << endl;
}
};
class Triangle
{
public:
Point a,b,c;
Triangle(){}
Triangle(Point x,Point y,Point z)
{
a =x;b=y;c=z;
double l1 = sqrt( (a.x-b.x)*(a.x-b.x)+(a.y-b.y)*(a.y-b.y) );
double l2 = sqrt( (a.x-c.x)*(a.x-c.x)+(a.y-c.y)*(a.y-c.y) );
double l3 = sqrt( (b.x-c.x)*(b.x-c.x)+(b.y-c.y)*(b.y-c.y) );
if(l1 +l2 <= l3 || l2+l3 <= l1 || l1+l3<= l2)
cout << "不能组成三角形";
}
double getzc()
{
double l1 = sqrt( (a.x-b.x)*(a.x-b.x)+(a.y-b.y)*(a.y-b.y) );
double l2 = sqrt( (a.x-c.x)*(a.x-c.x)+(a.y-c.y)*(a.y-c.y) );
double l3 = sqrt( (b.x-c.x)*(b.x-c.x)+(b.y-c.y)*(b.y-c.y) );
if(l1 +l2 <= l3 || l2+l3 <= l1 || l1+l3<= l2)
{
return -1;
}
else
return (l1 + l2 +l3);
}
};
int main()
{
Circle c(Point(2,2),1);
c.fun(Point(1,1));
Triangle tr(Point(0,0),Point(1,0),Point(0,1));
if (tr.getzc() < 0)
{
cout <<"不构成三角形"<<endl;
}else
cout <<"周长="<<tr.getzc()<<endl;
return 0;
}

``````
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参考如下：

``````#include <iostream>
#include <math.h>
using namespace std;

//抽象类
class MyObject
{
public:
virtual double getzc()=0;
};

class Point
{
public:
double x,y;
Point(){x=0;y=0;}
Point(int xx,int yy){x=xx;y=yy;}
};
class Circle :public MyObject
{
public:
Point center;
Circle(){}
Circle(Point p,double r)
{
center = p;
}
double getzc()
{
}
void fun(Point p)
{
double dis = sqrt( (center.x-p.x)*(center.x-p.x) + (center.y-p.y)*(center.y-p.y) );
cout << "在圆内"<< endl;
cout << "在圆上"<< endl;
else
cout << "在圆外" << endl;
}
};
class Triangle :public MyObject
{
public:
Point a,b,c;
Triangle(){}
Triangle(Point x,Point y,Point z)
{
a =x;b=y;c=z;
double l1 = sqrt( (a.x-b.x)*(a.x-b.x)+(a.y-b.y)*(a.y-b.y) );
double l2 = sqrt( (a.x-c.x)*(a.x-c.x)+(a.y-c.y)*(a.y-c.y) );
double l3 = sqrt( (b.x-c.x)*(b.x-c.x)+(b.y-c.y)*(b.y-c.y) );
if(l1 +l2 <= l3 || l2+l3 <= l1 || l1+l3<= l2)
cout << "不能组成三角形";
}
double getzc()
{
double l1 = sqrt( (a.x-b.x)*(a.x-b.x)+(a.y-b.y)*(a.y-b.y) );
double l2 = sqrt( (a.x-c.x)*(a.x-c.x)+(a.y-c.y)*(a.y-c.y) );
double l3 = sqrt( (b.x-c.x)*(b.x-c.x)+(b.y-c.y)*(b.y-c.y) );
if(l1 +l2 <= l3 || l2+l3 <= l1 || l1+l3<= l2)
{
return -1;
}
else
return (l1 + l2 +l3);
}
};
int main()
{
Circle c(Point(2,2),1);
c.fun(Point(1,1));
Triangle tr(Point(0,0),Point(1,0),Point(0,1));
if (tr.getzc() < 0)
{
cout <<"不构成三角形"<<endl;
}else
cout <<"周长="<<tr.getzc()<<endl;
return 0;
}

``````
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• 系统已结题 12月14日
• 已采纳回答 12月6日
• 创建了问题 12月5日

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