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题目：

设计一个简单的四人制扑克牌游戏，能够完成以下功能：

1. 洗牌
2. 发牌
3.自定义规则，在每轮单张出牌时，判定赢家
4.自定义规则，判定最终的赢家

规则简化版：

仅能出单张牌，且出牌时，每个人出的是自己手中牌中刚好能压过上家的最小牌，最先出完的为赢家

import randomfrom random import choiceflower = ['\u2660','\u2663','\u2665','\u2666']pai = ['3','4','5','6','7','8','9','10','J','Q','K','A','2']list = []list0 = []#储存发的牌list1 = []list2 = []list3 = []value = []value0 = []#储存牌代表的值value1 = []value2 = []value3 = []l00 = []#储存进行升序排序后的牌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)#随机选牌 for x in range(len(list)): if(k == list[x]):#和链表的牌进行匹配，删掉对应项 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))#将每个人的牌转换为字典形式d1 = dict(zip(list1,value1))d2 = dict(zip(list2,value2))d3 = dict(zip(list3,value3))l0 = sorted(d0.values())#对牌所代表的数字进行排序l1 = sorted(d1.values())l2 = sorted(d2.values())l3 = sorted(d3.values())#对发给每个人的牌进行排序for i in range(len(l0)): for j in list0: if(l0[i] == d0[j]): l00.append(j) breakfor i in range(len(l1)): for j in list1: if(l1[i] == d1[j]): l11.append(j) breakfor i in range(len(l2)): for j in list2: if(l2[i] == d2[j]): l22.append(j) breakfor 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("第一个人的牌：",l00)print("第二个人的牌：",l11)print("第三个人的牌：",l22)print("第四个人的牌：",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]):#判断是否当前牌中最大牌，若是，则该此人继续出牌 n = l0[0] l0.pop(0) break y = y + 1 if (len(l0) == 0): print("赢家：第一个人") 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("赢家：第二个人") 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("赢家：第三个人") 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("赢家：第四个人") break#将剩余牌从键值转化成牌if(len(l0) != 0): for i in range(len(l0)): for j in list0: if(l0[i] == d0[j]): l0[i] = j breakif(len(l1) != 0): for i in range(len(l1)): for j in list1: if(l1[i] == d1[j]): l1[i] = j breakif(len(l2) != 0): for i in range(len(l2)): for j in list2: if(l2[i] == d2[j]): l2[i] = j breakif(len(l3) != 0): for i in range(len(l3)): for j in list3: if(l3[i] == d3[j]): l3[i] = j breakprint("第一个人的牌：",l0)print("第二个人的牌：",l1)print("第三个人的牌：",l2)print("第四个人的牌：",l3)

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