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Python - 简化“石头剪刀”的逻辑条件

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我正在解决一个问题，它说：

在Big Bang Theory中，Sheldon和Raj创造了一款新游戏：“rock-paper-scissors-lizard-Spock” . 游戏规则是：剪刀剪纸;纸覆盖岩石;岩石碾碎蜥蜴;蜥蜴毒药斯波克; Spock砸碎剪刀;剪刀斩首蜥蜴;蜥蜴吃纸;论文反驳了斯波克; Spock蒸发岩石;岩石压碎剪刀 . 在Sheldon获胜的情况下，他会说：“Bazinga！”;如果Raj赢了，Sheldon会宣称：“Raj被骗了”;在关系中，他会要求一个新游戏：“再次！” . 考虑到两者选择的选项，制作一个程序，打印谢尔顿对结果的反应 . 输入包含一系列测试用例 . 第一行包含正整数T（T≤100），表示测试用例的数量 . 每个测试用例由输入行表示，分别包含Sheldon和Raj的选项，用空格分隔 .

我对这个问题的代码是

T = int(input())

for i in range(T):
    Sheldon, Raj = input().split(' ')

    if(Sheldon == "scissors" and (Raj == "paper" or Raj == "lizard")):
        Win = True
    elif(Sheldon == "lizard" and (Raj == "paper" or Raj == "Spock")):
        Win = True
    elif(Sheldon == "Spock" and (Raj == "rock" or Raj == "scissors")):
        Win = True
    elif(Sheldon == "paper" and (Raj == "rock" or Raj == "Spock")):
        Win = True
    elif(Sheldon == "rock" and (Raj == "scissors" or Raj == "lizard")):
        Win = True
    elif(Raj == "scissors" and (Sheldon == "paper" or Sheldon == "lizard")):
        Lose = True
    elif(Raj == "lizard" and (Sheldon == "paper" or Sheldon == "Spock")):
        Lose = True
    elif(Raj == "Spock" and (Sheldon == "rock" or Sheldon == "scissors")):
        Lose = True
    elif(Raj == "paper" and (Sheldon == "rock" or Sheldon == "Spock")):
        Lose = True
    elif(Raj == "rock" and (Sheldon == "scissors" or Sheldon == "lizard")):
        Lose = True
    elif(Sheldon == Raj):
        Tie = True

    if(Win == True):
        print("Case #{0}: Bazinga!".format(i+1))
    elif(Lose == True):
        print("Case #{0}: Raj cheated!".format(i+1))
    elif(Tie == True):
        print("Case #{0}: Again!".format(i+1))

    Win = Lose = Tie = False

但我觉得它太久了 . 有没有办法减少它？

python optimization

3 回答

  • 0

    首先，祝贺你试着写这篇文章！第一次尝试你的逻辑非常好 .

    下一步是制作一个数据结构，您可以按照相同的方式查询规则 . 一个很好的契合是 dictionary ：

    options = {
 'scissors': ('paper', 'lizard'),
 'paper': ('rock', 'spock'),
 'rock': ('lizard', 'scissors'),
 'lizard': ('spock', 'paper'),
 'spock': ('scissors', 'rock'),
}

    然后你可以查询它而不是重复大量的 if ：

    if raj == sheldon:
   print("Case #{0}: Again!".format(i+1))
elif raj in options[sheldon]:
   print("Case #{0}: Bazinga!".format(i+1))
else: 
   print("Case #{0}: Raj cheated!".format(i+1))
    回复于
  • 0
    T = int(input())

for i in range(T):
    Sheldon, Raj = input().split(' ')

    if(Sheldon == Raj):
        Tie = True
    elif((Sheldon == "scissors" and (Raj in ["paper","lizard"])) or
         (Sheldon == "lizard" and (Raj in ["paper","Spock"])) or
         (Sheldon == "Spock" and (Raj in ["rock","scissors"])) or
         (Sheldon == "paper" and (Raj in ["rock","Spock"])) or
         (Sheldon == "rock" and (Raj in ["scissors","lizard"]))
        ):
        Win = True
    else:
        Lose = True

    if(Win == True):
        print("Case #{0}: Bazinga!".format(i+1))
    elif(Lose == True):
        print("Case #{0}: Raj cheated!".format(i+1))
    elif(Tie == True):
        print("Case #{0}: Again!".format(i+1))

    Win = Lose = Tie = False
    回复于
  • 7

    试试这个，使用词典

    T = int(input())

for i in range(T):

    rules=  {
        "rock":     {"rock":0, "paper":-1,"scissors":1,"lizard":1,"Spock":-1},
        "paper":    {"rock":1, "paper":0,"scissors":-1,"lizard":-1,"Spock":1},
        "scissors": {"rock":-1, "paper":1,"scissors":0,"lizard":1,"Spock":-1},
        "lizard":   {"rock":1, "paper":-1,"scissors":1,"lizard":0,"Spock":-1},
        "Spock":    {"rock":1, "paper":-1,"scissors":1,"lizard":-1,"Spock":0}
        }
    Sheldon, Raj = input().split(' ') 

    Result = rules[Sheldon][Raj]
    if(Result == 1):
        print("Case #{0}: Bazinga!".format(i+1))
    elif(Result == -1):
        print("Case #{0}: Raj cheated!".format(i+1))
    else:
        print("Case #{0}: Again!".format(i+1))
    回复于

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