这里写自定义目录标题
链接:https://www.nowcoder.com/questionTerminal/f9c4290baed0406cbbe2c23dd687732c?toCommentId=3848030
来源:牛客网
在二维坐标系中,所有的值都是double类型,那么一个三角形可以由3个点来代表,给定3个点代表的三角形,再给定一个点(x, y),判断(x, y)是否在三角形中
输入描述:
输入有四行,每行两个浮点数。
前三行的6个数分别代表三角形的三个顶点的坐标
最后两个数分别表示(x, y)
输出描述:
若(x, y)在三角形中,输出"Yes"
否则输出"No"
示例1
输入
-1.00 0.00
1.50 3.50
2.73 -3.12
1.23 0.23
输出
Yes
python代码如下
写的比较冗余,希望可以帮到找工作的小伙伴
if __name__ == "__main__":
# 读取三个坐标点的值
p1=input().split()
p2=input().split()
p3=input().split()
p0=input().split()
"""
p1_x = float(input())
p1_y = float(input())
p2_x = float(input())
p2_y = float(input())
p3_x = float(input())
p3_y = float(input())
p0_x = float(input())
p0_y = float(input())
"""
p1_x = float(p1[0])
p1_y = float(p1[1])
p2_x = float(p2[0])
p2_y = float(p2[1])
p3_x = float(p3[0])
p3_y = float(p3[1])
p0_x = float(p0[0])
p0_y = float(p0[1])
p_x=[p1_x,p2_x,p3_x]
p_y=[p1_y,p2_y,p3_y]
if p0_x<min(p_x) or p0_x>max(p_x) or p0_y<min(p_y) or p0_y>max(p_y):
print("No")
else:
if p2_x == p1_x:
p02_y=((p3_y-p1_y)/(p3_x-p1_x))*(p0_x-p1_x)+p1_y
p01_y=((p2_y-p3_y)/(p2_x-p3_x))*(p0_x-p3_x)+p3_y
if p0_y==p01_y or p0_y==p02_y:
print(0)
elif p0_y>max(p01_y,p02_y) and p0_y<min(p01_y,p02_y):
print("No")
else:
print("Yes")
elif p2_x==p3_x:
p03_y=((p2_y-p1_y)/(p2_x-p1_x))*(p0_x-p1_x)+p1_y
p02_y=((p3_y-p1_y)/(p3_x-p1_x))*(p0_x-p1_x)+p1_y
if p0_y==p03_y or p0_y==p02_y:
print(0)
elif p0_y>max(p03_y,p02_y) and p0_y<min(p03_y,p02_y):
print("No")
else:
print("Yes")
elif p1_x==p3_x:
p03_y=((p2_y-p1_y)/(p2_x-p1_x))*(p0_x-p1_x)+p1_y
p01_y=((p2_y-p3_y)/(p2_x-p3_x))*(p0_x-p3_x)+p3_y
if p0_y==p01_y or p0_y==p03_y:
print(0)
elif p0_y>max(p01_y,p03_y) and p0_y<min(p01_y,p03_y):
print("No")
else:
print("Yes")
else:
p01_y=((p2_y-p3_y)/(p2_x-p3_x))*(p0_x-p3_x)+p3_y
p02_y=((p3_y-p1_y)/(p3_x-p1_x))*(p0_x-p1_x)+p1_y
p03_y=((p2_y-p1_y)/(p2_x-p1_x))*(p0_x-p1_x)+p1_y
p0123_y=[p01_y,p02_y,p03_y]
p0123_y.sort()
if min(p_y)<p0123_y[0]:
if p0_y>p0123_y[0] and p0_y<p0123_y[1]:
print("Yes")
elif p0_y in p0123_y:
print(0)
else:
print("No")
if max(p_y)>p0123_y[2]:
if p0_y<p0123_y[2] and p0_y>p0123_y[1]:
print("Yes")
elif p0_y in p0123_y:
print(0)
else:
print("No")