Python之"rock paper scissors"逻辑条件的简化

我正在解决一个问题，它说:

In Big Bang Theory, Sheldon and Raj created a new game: "rock-paper-scissors-lizard-Spock".

The rules of the game are:

• scissors cuts paper;
• paper covers rock;
• rock crushes lizard;
• lizard poisons Spock;
• Spock smashes scissors;
• scissors decapitates lizard;
• lizard eats paper;
• paper disproves Spock;
• Spock vaporizes rock;
• rock crushes scissors.

In the case of Sheldon's victory, he would've said: "Bazinga!"; if Raj had won, Sheldon would declare: "Raj cheated"; in ties, he would request a new game: "Again!". Given the options chosen by both, make a program that prints Sheldon reaction to the outcome.

The input consists of a series of test cases. The first line contains a positive integer T (T ≤ 100), which represents the number of test cases. Each test case is represented by a line of the input, containing the choices of Sheldon and Raj, respectively, separated by a space.

我这道题的代码是

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 

但我觉得它太长了。有什么办法可以减少吗？

请您参考如下方法:

首先，恭喜你尝试写这篇文章!对于第一次尝试，您的逻辑非常好。

下一步是制作一个数据结构，您可以用相同的方式查询规则。 字典 比较合适:

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 ifs:

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))