关于程序相关的
- 您可以反复玩游戏,直到选择停止为止。
- 该程序跟踪获胜情况。
- 大小写无关紧要(即ROCK与Rock相同)。
- 如果您输入的内容无效,程序会一直提示您,直到您输入有效的内容。
对项目进行编码的步骤:
- 创建一个简单的单轮游戏版本,我们不执行正确的输入。
- 如果输入了无效的内容,则添加while循环可重新提示用户输入选择。
- 使用while循环让用户反复播放,并使用变量来跟踪得分。
程序代码
import random input("Welcome to Rock, Paper, Scissors! Press Enter to start.") print() user_wins = 0 computer_wins = 0 choices = ["rock", "paper", "scissors"] while True: random_index = random.randint(0,2) cpu_choice = choices[random_index] user_choice = input("Rock, Paper, or Scissors? ").lower() while user_choice not in choices: user_choice = input("That is not a valid choice. Please try again: ").lower() print() print("Your choice:", user_choice) print("Computer's choice:", cpu_choice) print() if user_choice == 'rock': if cpu_choice == 'rock': print("It's a tie!") elif cpu_choice == 'scissors': print("You win!") user_wins+=1 elif cpu_choice == 'paper': print("You lose!") computer_wins+=1 elif user_choice == 'paper': if cpu_choice == 'paper': print("It's a tie!") elif cpu_choice == 'rock': print("You win!") user_wins+=1 elif cpu_choice == 'scissors': print("You lose!") computer_wins+=1 elif user_choice == 'scissors': if cpu_choice == 'scissors': print("It's a tie!") elif cpu_choice == 'paper': print("You win!") user_wins+=1 elif cpu_choice == 'rock': print("You lose!") computer_wins+=1 print() print("You have "+str(user_wins)+" wins") print("The computer has "+str(computer_wins)+" wins") print() repeat = input("Play again? (Y/N) ").lower() while repeat not in ['y', 'n']: repeat = input("That is not a valid choice. Please try again: ").lower() if repeat == 'n': break print("\n----------------------------\n")
运行效果:
以上就是Python制作简单的剪刀石头布游戏的详细内容,更多关于Python 剪刀石头布游戏的资料请关注自学编程网其它相关文章!
- 本文固定链接: https://zxbcw.cn/post/201912/
- 转载请注明:必须在正文中标注并保留原文链接
- QQ群: PHP高手阵营官方总群(344148542)
- QQ群: Yii2.0开发(304864863)