Question

我和我的朋友正在学校编写练习，我们遇到了如何让海龟保持正轨的困难。为了说明这个广场的第一面，两只乌龟正在两侧比赛。但是，当它们转90度时，其中1或2个会脱离侧面，这对于运动要求是不正确的。

这是我们的代码：

import turtle
from random import randint
def read_int(prompt,first,last):
    x = int(input(prompt))
    while x < first or x > last:
        print("Not in range. Try Again!!!")
        x= int(input(prompt))
    return x

square_count = read_int("Enter your laps between 1 and 10: ",1,10)
print(square_count)
#def t():

window = turtle.Screen()
window.bgcolor('lightblue')
def draw_square(turtle, center, size):
    xPt, yPt = center
    xPt -= size / 2
    yPt += size / 2
    side = 4
    size = 300
    angle = 90
    turtle.speed(0)
    turtle.up()
    turtle.goto(xPt, yPt)
    turtle.down()
    for i in range(side):
        turtle.forward(size)
        turtle.right(angle)

t = turtle.Turtle()
draw_square(t,(0,0),300)
t.shape('turtle')
t.color("red")
t.pensize(5)
t.up()
t.goto(-150, 150)

#def r():
r = turtle.Turtle()
draw_square(r,(0,0),300)
r.shape('turtle')
r.color("yellow")
r.up()
r.pensize(5)
r.goto(-150, 150)

sides = 4
size = 300


count_int = int(square_count)* sides
if count_int > 1:
    for sides in range(count_int):
            i = 0
            e = 0
            while i in range(0, size) or e in range(0, size):
                t_step = randint(1, 5)
                t.forward(t_step)
                i = i + t_step
                r_step = randint(1, 5)
                r.forward(r_step)
                e = e + r_step

            t.right(90)
            r.right(90)





window.exitonclick()

Answer 1

你的错误在于程序的结束。

while i in range(0, size) or e in range(0, size):
                t_step = randint(1, 5)
                t.forward(t_step)
                i = i + t_step
                r_step = randint(1, 5)
                r.forward(r_step)
                e = e + r_step

在假设i = 298的情况下，t_step选择一个随机整数为3.乌龟最终会移动大于300的范围，从而移动到绘制的框之外。 循环不会停止，因为它只会在运行所有代码时意识到i超出范围。（即，在环断裂时，乌龟已经移出了盒子。）这就是为什么乌龟出门了。您可以尝试使用“if”条件来阻止这种情况。

Answer 2

你有两种问题。
首先，你应该注意你的乌龟采取多少步骤。因为如果一个正方形的大小为300，那么乌龟可能会采取一定数量的步骤，其总和不等于300.我建议您将随机步骤的生成更改为：

steps = random.randrange(0, 4, 2)

通过这种方式，乌龟只需要0步或2步，你确定总和会带你到300.
第二点可能更重要。如果你像你一样构建种族循环，只有当另一只乌龟到达一条末尾时，乌龟才会转动，你应该重新组织你的代码，使两只乌龟独立于另一只。

Answer 3

我会采取略微不同的方法 - 而不是硬编码两只乌龟，设计任意数量的海龟，然后只比赛两只乌龟。这是基于任意数量的海龟的重写，包括一些样式代码的返工，以及@Claudio的一些代码建议：

from random import randint
from turtle import Turtle, Screen
from collections import defaultdict

SIZE = 300
SIDES = 4
ANGLE = 90
MAXIMUM_STRIDE = 5

COLORS = ('red', 'gold', 'green', 'orange', 'blue')

MAX_RACERS = len(COLORS)

def read_int(prompt, first, last):
    x = int(input(prompt))

    while not first <= x <= last:
        print('Not in range! Try Again.')
        x = int(input(prompt))

    return x

def draw_square(turtle, center, size):
    xPt, yPt = center

    xPt -= size / 2
    yPt += size / 2

    turtle.up()
    turtle.goto(xPt, yPt)
    turtle.down()

    for _ in range(SIDES):
        turtle.forward(SIZE)
        turtle.right(ANGLE)

lap_count = read_int('Enter number of laps (between 1 and 10): ', 1, 10)
no_racers = read_int('Enter number of racers (between 2 and {}): '.format(MAX_RACERS), 2, MAX_RACERS)

jockey_colors = COLORS[0:no_racers]

window = Screen()
window.bgcolor('lightblue')

racers = defaultdict(dict)

draw_square(Turtle(visible=False), (0, 0), SIZE)

for color in jockey_colors:
    jockey = Turtle('turtle', visible=False)
    jockey.speed('fastest')
    jockey.color(color)

    jockey.up()
    jockey.goto(-SIZE/2, SIZE/2)
    jockey.showturtle()

    racers[color]['jockey'] = jockey
    racers[color]['sides'] = 0
    racers[color]['position'] = 0

finished = False

while not finished:
    for racer in racers.values():
        jockey = racer['jockey']

        step = randint(1, MAXIMUM_STRIDE + 1)

        if racer['position'] + step > SIZE:

            racer['sides'] += 1

            if racer['sides'] == lap_count * SIDES:
                finished = True
                break

            baby_step = SIZE - racer['position']
            jockey.forward(baby_step)
            jockey.right(ANGLE)
            racer['position'] = 0

            step -= baby_step

        jockey.forward(step)
        racer['position'] += step

window.exitonclick()

Answer 4

你在这里：

import turtle

from random import randint

def read_int(prompt,first,last):
    x = int(input(prompt))
    while x < first or x > last:
        print("Not in range. Try Again!!!")
        x= int(input(prompt))
    return x

no_of_laps = read_int("Enter your laps between 1 and 10: ",1 , 10)
print(no_of_laps)

window = turtle.Screen()

window.bgcolor('lightblue')

def draw_square(turtle, center, size):
    xPt, yPt = center
    xPt -= size / 2
    yPt += size / 2
    side = 4
    size = 300
    angle = 90
    turtle.speed(0)
    turtle.up()
    turtle.goto(xPt, yPt)
    turtle.down()
    for i in range(side):
        turtle.forward(size)
        turtle.right(angle)

t = turtle.Turtle()
draw_square(t,(0,0),300)

t.shape('turtle')
t.color("red")
t.pensize(5)
t.up()
t.goto(-150, 150)

#def r():
r = turtle.Turtle()
draw_square(r,(0,0),300)
r.shape('turtle')
r.color("yellow")
r.up()
r.pensize(5)
r.goto(-150, 150)

sides = 4
size  = 300

no_of_turns = int(no_of_laps) * sides

countTurns_t = 0
countTurns_r = 0
t_i = 0
r_i = 0

while countTurns_t < no_of_turns and countTurns_r < no_of_turns: 


    t_step = randint(1, 5)
    if t_i + t_step > size:
        t_step = size - t_i
        t.forward(t_step)
        t_i = 0
        t.right(90)
        countTurns_t += 1
    else:
        t.forward(t_step)
        t_i += t_step

    r_step = randint(1, 5)
    if r_i + r_step > size:
        r_step = size - r_i
        r.forward(r_step)
        r_i = 0
        r.right(90)
        countTurns_r += 1
    else:
        r.forward(r_step)
        r_i += r_step

window.exitonclick()

P.S。希望您能看到上面提供的代码如何以及为什么按预期工作。仔细阅读其他答案，他们指出了编写代码时遇到的重要问题。

海龟赛车游戏 - 无法将乌龟留在广场

4 个答案: