不知道为什么纸牌游戏会使牌变差
不知道为什么纸牌游戏会使牌变差
提问于 2018-10-06 00:58:42
我正在制作一个名为“乞讨我的邻居”的纸牌游戏，在游戏中创建一副扑克牌，然后随机排序，然后平均分成2名玩家。然后，每个玩家抽签并把它放在桌子上，直到有一张惩罚牌(面牌)。每张牌面牌的债务值为1-4，而另一个牌手必须在牌桌上打这个数目的牌。然而，另一个玩家可以自己抽一张罚单，重新开始还债。如果一个玩家玩一个债务，而另一个玩家玩所有的牌而不是任何的债务，那么那些玩债务的玩家就会把所有的牌都拿走。胜利者是拥有所有牌牌的球员。
我的问题是，当游戏自行运行(控制台中的play())时，堆栈(每个玩家拥有的牌#)减少的不是1，而是一些任意的数量。我该怎么解决这个问题？
编辑：
if(G['debt']>0):
# Paying a debt.
print("Turn {}: Player {} is paying a debt.".format(turn, current(G)))
# May want to show debt cards being played with displayCard().
# Careful to handle any penalty cards paid as part of a debt!
if len(G['stacks'][G['next']]) > G['debt']:
#if
for i in range(G['debt']):
# Print what card is being played
print("Turn {}: Player {} Played {}".format(turn, current(G), (displayCard(G['stacks'][G['next']][0]))))
# Test if card being played is a penalty card
if G['stacks'][G['next']][0][0] == 1:
#G['stacks'][G['next']].pop(0)
G['debt']=4
i=0
elif G['stacks'][G['next']][0][0] == 13:
#G['stacks'][G['next']].pop(0)
G['debt']=3
i=0
elif G['stacks'][G['next']][0][0] == 12:
#G['stacks'][G['next']].pop(0)
G['debt']=2
i=0
elif G['stacks'][G['next']][0][0] == 11:
#G['stacks'][G['next']].pop(0)
G['debt']=1
i=0
# Add the card to the table
G['table'].append(G['stacks'][G['next']][0])
# Remove the card from the player's stack
G['stacks'][G['next']].pop()
else:
G['debt'] = 0原始代码：
from random import randint
def createDeck(N=13, S=('spades', 'hearts', 'clubs', 'diamonds')):
return([(v, s) for v in range(1,N+1) for s in S])
def displayCard(c):
suits = {'spades':'\u2660', 'hearts':'\u2661', 'diamonds':'\u2662', 'clubs':'\u2663'}
return(''.join( [ str(c[0]), suits[c[1]] ] ))
def simpleShuffle(D):
for i in range(len(D)):
r=randint(i,len(D)-1)
D[i],D[r]=D[r],D[i]
return(D)
def newGame(N=13, S=('spades', 'hearts', 'clubs', 'diamonds')):
d = simpleShuffle(createDeck(N,S))
return {'table':[], 'next':0, 'debt':0, 'stacks':[d[:len(d)//2],d[len(d)//2:]]}
def describeGame(G):
return('Player:'+str(G['next'])+' Stacks:['+str(len(G['stacks'][0]))+', '+str(len(G['stacks'][1]))+'] Table:'+str(len(G['table']))+' Debt:'+str(G['debt']))
def current(G):
return(G['next'])
def opponent(G):
if G['next']==0:
return(1)
else:
return(0)
def advancePlayer(G):
G['next']=opponent(G)
return(G)
def play(G=newGame()):
turn = 0
while(G['stacks'][0]!=0 and G['stacks'][1]!=0):
# Show the state of play.
print("Turn {}: {}".format(turn, describeGame(G)))
# Make a move. First, check to see if a debt is due. If so,
# pay it.
if(G['debt']>0):
# Paying a debt.
print("Turn {}: Player {} is paying a debt.".format(turn, current(G)))
if len(G['stacks'][G['next']]) >= G['debt']:
for i in range(G['debt']):
# Print what card is being played
print("Turn {}: Player {} Played {}".format(turn, current(G), (displayCard(G['stacks'][G['next']][0]))))
# Test if card being played is a penalty card
if G['stacks'][G['next']].pop(0) == 1:
G['debt']=4
i=0
elif G['stacks'][G['next']].pop(0) == 13:
G['debt']=3
i=0
elif G['stacks'][G['next']].pop(0) == 12:
G['debt']=2
i=0
elif G['stacks'][G['next']].pop(0) == 11:
G['debt']=1
i=0
# Add the card to the table
G['table'].append(G['stacks'][G['next']][0])
# Remove the card from the player's stack
G['stacks'][G['next']].pop(0)
# Increment turn
turn = turn + 1
else:
print("Turn {}: Player {} Played {}".format(turn, current(G), (displayCard(G['stacks'][G['next']][0]))))
#print(displayCard(G['stacks'][G['next']][0]))
# Check if c is a penalty card.
if(G['stacks'][G['next']][0][0]==1 or G['stacks'][G['next']][0][0]==11 or G['stacks'][G['next']][0][0]==12 or G['stacks'][G['next']][0][0]==13):
# Set up a new debt for the other player and advance
# immediately to next turn.
if (G['stacks'][G['next']][0][0])== 1:
G['debt']=4
elif (G['stacks'][G['next']][0][0])== 13:
G['debt']=3
elif (G['stacks'][G['next']][0][0])== 12:
G['debt']=2
else:
G['debt']=1
# Not a penalty card; add it to the table.
G['table'].append(G['stacks'][G['next']][0])
# Remove the card
G['stacks'][G['next']].pop(0)
# Advance to next player.
advancePlayer(G)
# Increment turn counter.
turn = turn + 1
# Exit loop: indicate winner.`enter code here`
print("Player {} wins in {} turns.".format(opponent(G), turn))回答 2
Stack Overflow用户
回答已采纳
发布于 2018-10-06 02:57:02
它停止了对我的窃听，并相应地对子弹进行计数。
def play(G=newGame()):
turn = 0
while(len(G['stacks'][0])!=0 and len(G['stacks'][1])!=0):
# Show the state of play.
print("Turn {}: {}".format(turn, describeGame(G)))
# Make a move. First, check to see if a debt is due. If so,
# pay it.
if(G['debt']>0):
# Paying a debt.
print("Turn {}: Player {} is paying a debt.".format(turn, current(G)))
if len(G['stacks'][G['next']]) >= G['debt']:
for i in range(G['debt']):
# Print what card is being played
print("Turn {}: Player {} Played {} for their debt.".format(turn, current(G), (displayCard(G['stacks'][G['next']][0]))))
nextcard = G['stacks'][G['next']][0][0]
# Test if card being played is a penalty card
if nextcard == 1:
G['debt']= 4
elif nextcard == 13:
G['debt']= 3
elif nextcard == 12:
G['debt']= 2
elif nextcard == 11:
G['debt']= 1
#G['stacks'][G['next']].pop(0)
# Add the card to the table
G['table'].append(G['stacks'][G['next']][0])
# Remove the card from the player's stack
G['stacks'][G['next']].pop(0)
# Increment turn
turn = turn + 1
# in each iteration the turn is increased
# however, outside of this loop the turn is increased once again
# take this into account and remove one turn
turn -= 1
else:
# player has less cards than they have to pay
# plays all his cards
print("Turn {}: Player {} has not enough cards to pay their debt.".format(turn, current(G)))
G['debt'] = 0
continue
else:
print("Turn {}: Player {} Played {}".format(turn, current(G), (displayCard(G['stacks'][G['next']][0]))))
#print(displayCard(G['stacks'][G['next']][0]))
# Check if c is a penalty card.
if(G['stacks'][G['next']][0][0]==1 or G['stacks'][G['next']][0][0]==11 or G['stacks'][G['next']][0][0]==12 or G['stacks'][G['next']][0][0]==13):
# Set up a new debt for the other player and advance
# immediately to next turn.
if (G['stacks'][G['next']][0][0])== 1:
G['debt']=4
elif (G['stacks'][G['next']][0][0])== 13:
G['debt']=3
elif (G['stacks'][G['next']][0][0])== 12:
G['debt']=2
else:
G['debt']=1
# Not a penalty card; add it to the table.
G['table'].append(G['stacks'][G['next']][0])
# Remove the card
G['stacks'][G['next']].pop(0)
# Advance to next player.
advancePlayer(G)
# Increment turn counter.
turn = turn + 1
# Exit loop: indicate winner.`enter code here`
print("Player {} wins in {} turns.".format(opponent(G), turn))我做的第一件事是将while条件更改为len()，而不仅仅是堆栈。我还引入了一个变量nextcard：
nextcard = G['stacks'][G['next']][0][0]我实现了您提到的其他条件(将债务设置为0)。
Stack Overflow用户
发布于 2018-10-06 01:12:03
我认为问题就在这个代码中：
if G['stacks'][G['next']].pop(0) == 1:
G['debt']=4
i=0
elif G['stacks'][G['next']].pop(0) == 13:
G['debt']=3
i=0
elif G['stacks'][G['next']].pop(0) == 12:
G['debt']=2
i=0
elif G['stacks'][G['next']].pop(0) == 11:
G['debt']=1
i=0当一张卡片被弹出时，它被从堆栈中移除。每次执行“检查”时都会这样做。因此，当check返回true时，任意数字来自。如果是的话。13，它将移除2张卡片，价值为12，它将移除3张卡片，等等。
你不应该打开卡片，你应该检查它，就像你以后做的那样：
if G['stacks'][G['next']][0][0])== 1:页面原文内容由Stack Overflow提供。腾讯云小微IT领域专用引擎提供翻译支持
原文链接：
https://stackoverflow.com/questions/52674825
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