Python - 简化"石头剪刀"的逻辑条件

Question

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？

Answer 1

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))