岩石剪刀紙程序

Question

這是用python編寫岩石剪刀紙遊戲的代碼。 如果我運行的代碼，它的作品，但是，當它成爲領帶時，它會像這樣輸出。 無論如何，當我得到領帶的結果時，我可以消除打印（圓）嗎？ 我想看看它如在示例底部所示

********************* ROUND＃1 ********* ************

挑選你的投球：[ock，[p] aper或[s] cissors？ p 扎！

*********************＃1 *********************

挑選你的投球：[ock，[p] aper或[s] cissors？ s 電腦扔石頭，你輸了！

你的分數：0 電腦積分：1

********************* ROUND＃3 ********* ************

挑選你的投球：[ock，[p] aper或[s] cissors？ s 扎！

選擇你的選擇：[r] ock，[p] aper或[s] cissors？ p 扎！

選擇你的選擇：[r] ock，[p] aper或[s] cissors？ r 電腦扔剪刀，你贏了！

你的分數：2 電腦的得分：1

# A Python program for the Rock, Paper, Scissors game. 
import random 

def rock_paper_scissors(): 
''' Write your code for playing Rock Paper Scissors here. ''' 
user = 0 
computer = 0 
rounds = 1 

print() 
score = (int(input('How many points does it take to win? '))) 
print() 

while (computer < score and user < score): 
    RPS = random.randint(0,2) 
    if (RPS == 0): 
     RPS = 'rock' 
    elif (RPS == 1): 
     RPS = 'paper' 
    elif(RPS == 2): 
     RPS = 'scissors' 


    print('*'*21 + 'ROUND #'+str(rounds) + '*'*21) 
    print() 
    player = (input('Pick your throw: [r]ock, [p]aper, or [s]cissors? ')) 


    if RPS == 'rock' and player == 'r': 
     print('Tie!') 

    elif RPS == 'rock' and player == 's': 
     print('Computer threws rock, you lose!') 
     computer+=1 
     rounds += 1 
     print() 
     print('Your Score: ',user) 
     print('Computer Score: ',computer) 
    elif RPS == 'rock' and player == 'p': 
     print('Computer threw rock, you win!') 
     user+=1 
     rounds +=1 
     print() 
     print('Your Score: ',user) 
     print('Computer Score: ',computer)    
    if RPS == 'paper' and player == 'p': 
     print('Tie!') 
    elif RPS == 'paper' and player == 'r': 
     print('Computer threw paper, you lose!') 
     computer +=1 
     rounds += 1 
     print() 
     print('Your Score: ',user) 
     print('Computer Score: ',computer)    
    elif RPS == 'paper' and player == 's': 
     print('Computer threw paper, you win!') 
     user +=1 
     rounds +=1 
     print() 
     print('Your Score: ',user) 
     print('Computer Score: ',computer)   
    if RPS == 'scissors' and player == 's': 
     print('Tie!') 
    elif RPS == 'scissors'and player == 'p': 
     print('Computer threw scissors, you lose!') 
     computer +=1 
     rounds+=1 
     print() 
     print('Your Score: ',user) 
     print('Computer Score: ',computer)    
    elif RPS == 'scissors' and player == 'r': 
     print('Computer threw scissors, you win!') 
     user +=1 
     rounds+=1 
     print() 
     print('Your Score: ',user) 
     print('Computer Score: ',computer)    
    print() 

if user> computer: 
    print('*'*21 + 'GAME OVER' + '*'*21) 
    print('You win!') 
else: 
    print('*'*21 + 'GAME OVER' + '*'*21) 
    print('Computer win!') 

print()   

def main(): 
print('ROCK PAPER SCISSORS in Python') 
print() 
print('Rules: 1) Rock wins over Scissors.') 
print('  2) Scissors wins over Paper.') 
print('  3) Paper wins over Rock.') 

rock_paper_scissors() 

main() 

Answer 1

我建議創建一個布爾was_tied，將其設置在程序的開始等於False，並在將其設置爲任何True或False每個可能結果的結束。然後，您可以將您的打印輪代碼放入if not was_tied聲明中。

Answer 2

所以我認爲你有一些是可以簡化這個代碼領域，但對於一個快速的解決方案，在這一行：

print('*'*21 + 'ROUND #'+str(rounds) + '*'*21) 
print() 

將其更改爲：

if not previous_round_was_tie: 
    print('*'*21 + 'ROUND #'+str(rounds) + '*'*21) 
    print() 

而在所有如果您的領帶方案爲真，請添加一行previous_round_was_tie = True。您還必須創建布爾值並將其設置爲您的while循環外部的False。