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python代码实现人机压指游戏
游戏规则:拇指赢食指,食指赢中指,中指赢无名指,无名指赢小指,小指赢拇指,不相邻手指平局。
游戏要求:计算机随机选择手指,用户键盘输入并控制游戏结束,赢一局得分+1,平局得分+0,输一局得分-1,统计输赢
欢迎大家进一步优化代码,谢谢!
- # python代码实现人机压指游戏
- # 游戏规则:拇指赢食指,食指赢中指,中指赢无名指,无名指赢小指,小指赢拇指,不相邻手指平局。
- # 游戏要求:计算机随机选择手指,用户键盘输入并控制游戏结束,赢一局得分+1,平局得分+0,输一局得分-1,统计输赢
- import random
- def play_game():
- # 定义一个字典,用来存储指针的名称
- options = {1: "拇指", 2: "食指", 3: "中指", 4: "无名指", 5: "小指"}
- # 定义一个字典,用来存储分数
- scores = {"胜": 1, "平": 0, "负": -1}
- # 定义用户分数
- user_score = 0
- # 定义电脑分数
- computer_score = 0
- # 定义计数器
- count = 0
- # 定义平局计数器
- count_0 = 0
- while True:
- # 电脑随机出指针
- computer_choice = random.randint(1, 5)
- print(f"计算机出了{options[computer_choice]}", computer_choice) # 人为作弊
- # 用户出指针
- user_choice = int(input("请选择手指(拇指:1,食指:2,中指:3,无名指:4,小指:5):"))
- user_finger = options[user_choice]
- computer_finger = options[computer_choice]
- # 判断用户和电脑出的指针是否满足条件
- if (user_choice == 1 and computer_choice == 2) or \
- (user_choice == 2 and computer_choice == 3) or \
- (user_choice == 3 and computer_choice == 4) or \
- (user_choice == 4 and computer_choice == 5) or \
- (user_choice == 5 and computer_choice == 1):
- # 满足条件,用户获胜
- result = "胜"
- user_score += 1
- elif computer_choice - user_choice == -1 or computer_choice - user_choice == 4:
- # 满足条件,电脑获胜
- result = "负"
- computer_score += 1
- else:
- # 不满足条件,平局
- result = "平局"
- user_score += 0
- computer_score += 0
- count_0 += 1
- count += 1
- print("你选择了:%s,计算机选择了:%s,结果:%s" % (user_finger, computer_finger, result))
- # 判断是否继续游戏
- continue_game = input("是否继续游戏?(Y/N)").upper()
- if continue_game != "Y":
- break
- print("游戏结束")
- print(f"总共玩了{count}局:")
- print(f'平局次数:{count_0}')
- print("你的得分:%d,计算机得分:%d" % (user_score, computer_score))
- if user_score > computer_score:
- print("恭喜:你赢了!!!")
- elif user_score < computer_score:
- print("抱歉:你输了!!!")
- else:
- print("平局!")
- play_game()
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