Question

下面有一段代码，其中随机选择了三个数字，用户有10次尝试来解决代码。我想添加一个作弊代码（可以这么说），以允许用户继续前进到下一个类/函数，而不必猜测三位数。

我尝试在or guess == 'cheat'之后添加while guess != code and guesses < 10。我尝试将if guess == code:转换为elif guess == code并在elif之前添加if guess == 'cheat'。我是一个菜鸟，我正在学习的来源说我的目标可以在一行中用两个单词完成。我迷路了... ...

 code = "%d%d%d" % (randint(1,9), randint(1,9), randint(1,9))
 guess = raw_input("[keypad]> ")
 guesses = 0

 while guess != code and guesses < 9:
        print "WRONG!"
        guesses += 1
        guess = raw_input("Guess 3 numbers> ")

    if guess == code:
        print "Correct!"
        return my_func()
    else:
        print "Game over"
        return exit()

Answer 1

首先，确保您的代码正确缩进。由于你有它的工作，它似乎只是问题上的格式问题。

有点不清楚这个作弊码会是什么，但我会把它作为两件事之一：作弊或其他字符串。无论哪种方式，它都可以存储在名为cheat的变量中，当猜测为code或cheat时，您可以退出while循环。

cheat = 'cheat'

while guess not in [code, cheat]:

还不清楚使用cheat时会发生什么，有两种可能的情况：

cheat以失败告终。

code = "%d%d%d" % (randint(1,9), randint(1,9), randint(1,9))
cheat = 'cheat'
guess = raw_input("[keypad]> ")
guesses = 0

while guess not in [cheat, code] and guesses < 9:
    print "WRONG!"
    guesses += 1
    guess = raw_input("Guess 3 numbers> ")

if guess == code:
    print "Correct!"
    return my_func()
else:
    print "Game over"
    return exit()

cheat以胜利结束游戏。

code = "%d%d%d" % (randint(1,9), randint(1,9), randint(1,9))
cheat = 'cheat'
guess = raw_input("[keypad]> ")
guesses = 0

while guess not in [cheat, code] and guesses < 9:
    print "WRONG!"
    guesses += 1
    guess = raw_input("Guess 3 numbers> ")

if guess in [cheat, code]:
    print "Correct!"
    return my_func()
else:
    print "Game over"
    return exit()

如何添加“作弊代码”来猜测我拥有的数字代码？

1 个答案: