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五十九、石头剪刀布

原文：http://inventwithpython.com/bigbookpython/project59.html

在这个版本的双人手游中，玩家面对电脑。你可以选择石头、布或剪刀。石头打败剪刀，剪刀打败布，布打败石头。这个程序增加了一些短暂的停顿来制造悬念。

这个游戏的一个变种，见项目 60，“石头剪刀布（必胜版本）。”

运行示例

当您运行rockpaperscissors.py时，输出将如下所示：

Rock, Paper, Scissors, by Al Sweigart email@protected
- Rock beats scissors.
- Paper beats rocks.
- Scissors beats paper.

0 Wins, 0 Losses, 0 Ties
Enter your move: (R)ock (P)aper (S)cissors or (Q)uit
> r
ROCK versus...
1...
2...
3...
SCISSORS
You win!
1 Wins, 0 Losses, 0 Ties
Enter your move: (R)ock (P)aper (S)cissors or (Q)uit
`--snip--`

工作原理

石头剪刀布的游戏逻辑相当简单，我们在这里用if - elif语句实现它。为了增加一点悬念，第 45 至 51 行在揭示对手的移动之前倒计时，在计数之间有短暂的停顿。这给了玩家一段时间，让他们对游戏的结果感到兴奋。如果没有这种停顿，结果会在玩家开始移动时立即出现——有点虎头蛇尾。为玩家改善用户体验并不需要很多代码。

"""Rock, Paper, Scissors, by Al Sweigart email@protected
The classic hand game of luck.
This code is available at https://nostarch.com/big-book-small-python-programming
Tags: short, game"""

import random, time, sys

print('''Rock, Paper, Scissors, by Al Sweigart email@protected
- Rock beats scissors.
- Paper beats rocks.
- Scissors beats paper.
''')

# These variables keep track of the number of wins, losses, and ties.
wins = 0
losses = 0
ties = 0

while True:  # Main game loop.
    while True:  # Keep asking until player enters R, P, S, or Q.
        print('{} Wins, {} Losses, {} Ties'.format(wins, losses, ties))
        print('Enter your move: (R)ock (P)aper (S)cissors or (Q)uit')
        playerMove = input('> ').upper()
        if playerMove == 'Q':
            print('Thanks for playing!')
            sys.exit()

        if playerMove == 'R' or playerMove == 'P' or playerMove == 'S':
            break
        else:
            print('Type one of R, P, S, or Q.')

    # Display what the player chose:
    if playerMove == 'R':
        print('ROCK versus...')
        playerMove = 'ROCK'
    elif playerMove == 'P':
        print('PAPER versus...')
        playerMove = 'PAPER'
    elif playerMove == 'S':
        print('SCISSORS versus...')
        playerMove = 'SCISSORS'

    # Count to three with dramatic pauses:
    time.sleep(0.5)
    print('1...')
    time.sleep(0.25)
    print('2...')
    time.sleep(0.25)
    print('3...')
    time.sleep(0.25)

    # Display what the computer chose:
    randomNumber = random.randint(1, 3)
    if randomNumber == 1:
        computerMove = 'ROCK'
    elif randomNumber == 2:
        computerMove = 'PAPER'
    elif randomNumber == 3:
        computerMove = 'SCISSORS'
    print(computerMove)
    time.sleep(0.5)

    # Display and record the win/loss/tie:
    if playerMove == computerMove:
        print('It\'s a tie!')
        ties = ties + 1
    elif playerMove == 'ROCK' and computerMove == 'SCISSORS':
        print('You win!')
        wins = wins + 1
    elif playerMove == 'PAPER' and computerMove == 'ROCK':
        print('You win!')
        wins = wins + 1
    elif playerMove == 'SCISSORS' and computerMove == 'PAPER':
        print('You win!')
        wins = wins + 1
    elif playerMove == 'ROCK' and computerMove == 'PAPER':
        print('You lose!')
        losses = losses + 1
    elif playerMove == 'PAPER' and computerMove == 'SCISSORS':
        print('You lose!')
        losses = losses + 1
    elif playerMove == 'SCISSORS' and computerMove == 'ROCK':
        print('You lose!')
        losses = losses + 1

在输入源代码并运行几次之后，尝试对其进行实验性的修改。你也可以自己想办法做到以下几点：

  • 在游戏中加入“蜥蜴”和“斯波克”的招式。蜥蜴毒死斯波克，吃纸，却被石头碾碎，被剪刀斩首。斯波克折断剪刀，蒸发岩石，但被蜥蜴毒死，被纸证明是错误的。
  • 允许玩家每次胜利赢得一分，每次失败失去一分。一旦获胜，玩家还可以承担“双倍或零”的风险，以可能在随后的回合中赢得 2、4、8、16 以及更多的点数。

探索程序

试着找出下列问题的答案。尝试对代码进行一些修改，然后重新运行程序，看看这些修改有什么影响。

  1. 如果将第 54 行的random.randint(1, 3)改为random.randint(1, 300)，会得到什么错误？
  2. 如果把第 65 行的playerMove == computerMove改成True会怎么样？