Python猜谜游戏不完全鳕鱼

Question

有人能帮我重新设计这段代码，让程序提示用户选择简单，中等还是困难。在

Easy: maxNumber = 10
Medium: maxNumber = 50
Hard: maxNumber = 100

它应该选择一个介于0和maxNumber之间的随机数。 该程序将循环调用一个函数来让用户猜测，然后调用另一个函数来检查他们的猜测。一个名为“getGuess”的函数，它将要求用户进行猜测和重新编译 如果猜测不在0和maxNumber之间 名为“checkGuess”的r函数将检查用户的猜测 与答案相比。 如果数字更高，函数将返回“更高” 如果数字小于猜测值，则为“低”；如果数字等于猜测值，则为“正确”。 一旦用户猜对了数字，程序将显示他们所有的猜测和 他们猜了多少次。然后程序会询问用户是否愿意尝试 再次显示难度菜单。在

^{pr2}$

Answer 1

你可以这样做：

from random import randint
myName = input("what's your name? ")
def pre_game():
    difficulty = input("Choose difficulty: type easy medium or hard: ")
    main_loop(difficulty)

def main_loop(difficulty):
    if difficulty == "easy":
        answer = randint(0, 10)
    elif difficulty == "medium":
        answer = randint(0, 50)
    else:
        answer = randint(0, 100)
    times_guessed = 0
    guess = int()
    while times_guessed < 6:
        print('Take a guess.')
        guess = input()
        guess = int(guess)

        times_guessed += 1
        if guess < answer:
            print('Your guess is too low.')

        if guess > answer:
            print('Your guess is too high.')

        if guess == answer:
            break


    if guess == answer:
        guessesTaken = str(times_guessed)
        print('Good job, ' + myName + '! You guessed my number in ' + guessesTaken + ' guesses!')

    if guess != answer:
        print('Nope. The number I was thinking of was ' + str(answer))
    next = input("Play again? y/n: ")
    if next == "y":
        pre_game()
    else:
        print("Thanks for playing!")
pre_game()