python如何实现四人制扑克牌游戏-创新互联
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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 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 = []#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) 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 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)
名称栏目:python如何实现四人制扑克牌游戏-创新互联
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