python编写函数判断三角形_任意三角形形状判断|Python练习系列[10]

完整代码和注释如下

class point(object):#定义平面点类

"""docstring for point"""

def __init__(self,x,y,name):

self.x = x

self.y = y

self.name = name

def distance(self,p2):#两点距离公式

self.d=((self.x-p2.x)**2+(self.y-p2.y)**2)**0.5

return self.d

def getd(self,p2):#获取两点距离

self.distance(p2)

print('The distance of ({},{}) and ({},{}) is {} '.format(self.x,self.y,p2.x,p2.y,self.d))

def istriangle(self,p2,p3):#判断这三点能否形成一个三角形

self.l_list=[]#先获取这三点构成的三条线段的长度

self.l_list.append(self.distance(p3))

self.l_list.append(p2.distance(p3))

self.l_list.append(self.distance(p2))

self.l_list.sort()#线段长度由小到大排序

if (self.l_list[0]+self.l_list[1]>self.l_list[2]) and (self.l_list[1]+self.l_list[2]>self.l_list[0]) and (self.l_list[2]+self.l_list[0]>self.l_list[1]):#长度判断

return 'can'

else:

return 'can not'

def getresult(self,p2,p3):#获取能否形成三角形的结果

result=self.istriangle(p2,p3)

print('Points:',self.name,p2.name,p3.name,result,'form a triangle')

def whichtriangle(self,p2,p3):#判断是哪种三角形

result=self.istriangle(p2,p3)

if result=='can not':

return print('Points:',self.name,p2.name,p3.name,result,'form a triangle')

if self.l_list[0]**2+self.l_list[1]**2>self.l_list[2]**2:#锐角

print('Points:',self.name,p2.name,p3.name,result,'form a acute triangle')

elif self.l_list[0]**2+self.l_list[1]**2==self.l_list[2]**2:#直角

print('Points:',self.name,p2.name,p3.name,result,'form a right triangle')

elif self.l_list[0]**2+self.l_list[1]**2

print('Points:',self.name,p2.name,p3.name,result,'form a obtuse triangle')

#测试

p1=point(12,-5,'p1')

p2=point(16,18,'p2')

p3=point(9,7,'p3')

p1.getd(p2)

p1.getresult(p2,p3)

p1.whichtriangle(p2,p3)

print()

p5=point(0,3,'p5')

p6=point(4,0,'p6')

p7=point(0,0,'p7')

p5.getd(p6)

p5.whichtriangle(p6,p7)

print()

p8=point(10,3,'p8')

p9=point(4,3,'p9')

p10=point(-9,0,'p10')

p8.getd(p9)

p8.whichtriangle(p9,p10)