• 函数递归 案例一 分形树


    # 绘制分形树
    import turtle


    def draw_brach(brach_length):

    if brach_length > 5:
    # 绘制右侧树枝
    turtle.pencolor("brown")
    turtle.pensize(3)
    turtle.forward(brach_length)
    turtle.right(30)
    draw_brach(brach_length - 15)

    turtle.left(60)
    draw_brach(brach_length - 15)
    # 绘制左侧树枝
    turtle.right(30)
    turtle.up()
    turtle.backward(brach_length)
    turtle.down()

    if brach_length<=10:
    turtle.pencolor("green")
    turtle.pensize(5)
    turtle.forward(brach_length)
    turtle.right(30)
    draw_brach(brach_length - 15)

    turtle.left(60)
    draw_brach(brach_length - 15)

    turtle.right(30)
    turtle.up()
    turtle.backward(brach_length)
    turtle.down()


    def main():
    turtle.speed(100)
    turtle.left(90)
    turtle.up()
    turtle.backward(200)
    turtle.down()
    turtle.pensize(10)
    draw_brach(100)

    turtle.exitonclick()
    if __name__ == '__main__':
    main()

  • 相关阅读:
    seajs快速了解
    lazyload.js详解
    iScroll-js—“smooth scrolling for the web”
    Backbone学习笔记一Backbone中的MVC
    JMH基准测试框架
    idea 下运行安卓项目
    安卓
    C++
    看完
    四叉树的js实现
  • 原文地址:https://www.cnblogs.com/kwkk978113/p/13308790.html
Copyright © 2020-2023  润新知