python實現四人制撲克牌游戲
更新時間:2020年04月22日 11:29:42 作者:dataocean
這篇文章主要介紹了python實現四人制撲克牌游戲,文中示例代碼介紹的非常詳細,具有一定的參考價值,感興趣的小伙伴們可以參考一下
本文實例為大家分享了python實現四人制撲克牌游戲的具體代碼,供大家參考,具體內容如下
題目:
設計一個簡單的四人制撲克牌游戲,能夠完成以下功能:
1. 洗牌
2. 發(fā)牌
3.自定義規(guī)則,在每輪單張出牌時,判定贏家
4.自定義規(guī)則,判定最終的贏家
規(guī)則簡化版:
僅能出單張牌,且出牌時,每個人出的是自己手中牌中剛好能壓過上家的最小牌,最先出完的為贏家
import random
from random import choice
flower = ['\u2660','\u2663','\u2665','\u2666']
pai = ['3','4','5','6','7','8','9','10','J','Q','K','A','2']
list = []
list0 = []#儲存發(fā)的牌
list1 = []
list2 = []
list3 = []
value = []
value0 = []#儲存牌代表的值
value1 = []
value2 = []
value3 = []
l00 = []#儲存進行升序排序后的牌
l11 = []
l22 = []
l33 = []
for i in flower:
for j in pai:
list.append(i+j)
for i in range(4):
for j in range(13):
value.append(j)
d = dict(zip(list,value))
for i in range(13):
for j in range(4):
if(j == 0):
k = choice(list)#隨機選牌
for x in range(len(list)):
if(k == list[x]):#和鏈表的牌進行匹配,刪掉對應項
value0.append(d[k])
list.pop(x)
break
list0.append(k)
if(j == 1):
k = choice(list)
for x in range(len(list)):
if (k == list[x]):
value1.append(d[k])
list.pop(x)
break
list1.append(k)
if (j == 2):
k = choice(list)
for x in range(len(list)):
if (k == list[x]):
value2.append(d[k])
list.pop(x)
break
list2.append(k)
if (j == 3):
k = choice(list)
for x in range(len(list)):
if (k == list[x]):
value3.append(d[k])
list.pop(x)
break
list3.append(k)
d0 = dict(zip(list0,value0))#將每個人的牌轉換為字典形式
d1 = dict(zip(list1,value1))
d2 = dict(zip(list2,value2))
d3 = dict(zip(list3,value3))
l0 = sorted(d0.values())#對牌所代表的數字進行排序
l1 = sorted(d1.values())
l2 = sorted(d2.values())
l3 = sorted(d3.values())
#對發(fā)給每個人的牌進行排序
for i in range(len(l0)):
for j in list0:
if(l0[i] == d0[j]):
l00.append(j)
break
for i in range(len(l1)):
for j in list1:
if(l1[i] == d1[j]):
l11.append(j)
break
for i in range(len(l2)):
for j in list2:
if(l2[i] == d2[j]):
l22.append(j)
break
for i in range(len(l0)):
for j in list3:
if(l3[i] == d3[j]):
l33.append(j)
break
# y = choice(['0','1','2','3'])
print("第一個人的牌:",l00)
print("第二個人的牌:",l11)
print("第三個人的牌:",l22)
print("第四個人的牌:",l33)
y = random.randint(0,3)
if (y == 0):
y = y + 1
n = l0[0]
l0.pop(0)
elif (y == 1):
y = y + 1
n = l1[0]
l1.pop(0)
elif (y == 2):
y = y + 1
n = l2[0]
l2.pop(0)
elif (y == 3):
y = 0
n = l3[0]
l3.pop(0)
for i in range(13):
if(y == 0):
for j in range(len(l0)):
if(l0[j] > n):
n = l0[j]
l0.pop(j)
if(len(l0) != 0 and n >= l1[len(l1)-1] and n >= l2[len(l2)-1] and n >= l3[len(l3)-1]):#判斷是否當前牌中最大牌,若是,則該此人繼續(xù)出牌
n = l0[0]
l0.pop(0)
break
y = y + 1
if (len(l0) == 0):
print("贏家:第一個人")
break
if(y == 1):
for j in range(len(l1)):
if(l1[j] > n):
n = l1[j]
l1.pop(j)
if (len(l1) != 0 and n >= l0[len(l0) - 1] and n >= l2[len(l2) - 1] and n >= l3[len(l3) - 1]):
n = l1[0]
l1.pop(0)
break
y = y + 1
if (len(l1) == 0):
print("贏家:第二個人")
break
if(y == 2):
for j in range(len(l2)):
if(l2[j] > n):
n = l2[j]
l2.pop(j)
if (len(l2) != 0 and n >= l0[len(l0) - 1] and n >= l1[len(l1) - 1] and n >= l3[len(l3) - 1]):
n = l2[0]
l2.pop(0)
break
y = y + 1
if (len(l2) == 0):
print("贏家:第三個人")
break
if (y == 3):
for j in range(len(l3)):
if (l3[j] > n):
n = l3[j]
l3.pop(j)
if (len(l3) != 0 and n >= l0[len(l0) - 1] and n >= l1[len(l1) - 1] and n >= l2[len(l2) - 1]):
n = l3[0]
l3.pop(0)
break
y = 0
if (len(l3) == 0):
print("贏家:第四個人")
break
#將剩余牌從鍵值轉化成牌
if(len(l0) != 0):
for i in range(len(l0)):
for j in list0:
if(l0[i] == d0[j]):
l0[i] = j
break
if(len(l1) != 0):
for i in range(len(l1)):
for j in list1:
if(l1[i] == d1[j]):
l1[i] = j
break
if(len(l2) != 0):
for i in range(len(l2)):
for j in list2:
if(l2[i] == d2[j]):
l2[i] = j
break
if(len(l3) != 0):
for i in range(len(l3)):
for j in list3:
if(l3[i] == d3[j]):
l3[i] = j
break
print("第一個人的牌:",l0)
print("第二個人的牌:",l1)
print("第三個人的牌:",l2)
print("第四個人的牌:",l3)
以上就是本文的全部內容,希望對大家的學習有所幫助,也希望大家多多支持腳本之家。

