帮我写一个21点游戏,用python实现。
时间: 2023-05-12 20:03:20 浏览: 61
好的,我可以为您提供一个简单的21点游戏的Python实现,代码如下:
```python
import random
def deal_card():
"""发一张牌"""
cards = [2, 3, 4, 5, 6, 7, 8, 9, 10, 10, 10, 10, 11]
return random.choice(cards)
def calculate_score(cards):
"""计算牌面分数"""
score = sum(cards)
if score > 21 and 11 in cards:
cards.remove(11)
cards.append(1)
score = sum(cards)
return score
def compare(user_score, computer_score):
"""比较分数"""
if user_score > 21 and computer_score > 21:
return "你输了,电脑赢了!"
elif user_score == computer_score:
return "平局!"
elif computer_score == 21:
return "你输了,电脑赢了!"
elif user_score == 21:
return "你赢了,恭喜!"
elif user_score > 21:
return "你输了,电脑赢了!"
elif computer_score > 21:
return "你赢了,恭喜!"
elif user_score > computer_score:
return "你赢了,恭喜!"
else:
return "你输了,电脑赢了!"
def play_game():
"""玩游戏"""
user_cards = []
computer_cards = []
game_over = False
for _ in range(2):
user_cards.append(deal_card())
computer_cards.append(deal_card())
while not game_over:
user_score = calculate_score(user_cards)
computer_score = calculate_score(computer_cards)
print(f"你的牌面:{user_cards},当前分数:{user_score}")
print(f"电脑的牌面:[{computer_cards[0]}, *]")
if user_score == 21 or computer_score == 21 or user_score > 21:
game_over = True
else:
draw_another_card = input("是否要再要一张牌?(y/n)").lower()
if draw_another_card == "y":
user_cards.append(deal_card())
else:
game_over = True
while computer_score < 17 and computer_score < user_score:
computer_cards.append(deal_card())
computer_score = calculate_score(computer_cards)
print(f"你的最终牌面:{user_cards},最终分数:{user_score}")
print(f"电脑的最终牌面:{computer_cards},最终分数:{computer_score}")
print(compare(user_score, computer_score))
play_game()
```
希望这个游戏能够让您玩得愉快!
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