C语言自动发牌程序，供大家参考，具体内容如下
一副扑克有52张牌，打桥牌时应将牌分给4个人。请设计一个程序完成自动发牌的工作。要求：黑桃用S (Spaces)表示，红桃用H (Hearts)表示，方块用D (Diamonds)表示，梅花用C (Clubs)表示。
分析：
要设置数组表现扑克牌
要设置数组表现玩家
要给扑克牌做特定标识，得到结果后玩家要知道自己手中黑桃有哪些、方块有哪些
初步想法：
设置4个字符数组保存4种梅花牌，设置4个字符数组表示4名玩家分配到的牌
每张牌随机发给4名玩家，当玩家的持牌数达到13，不再分配给该名玩家牌
代码展示：
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void mycode_13(){ srand(unsigned(time(NULL))); /*全部牌*/ char S[13] = { '2', '3', '4', '5', '6', '7', '8', '9', 'T', 'J', 'Q', 'K', 'A' }; char H[13] = { '2', '3', '4', '5', '6', '7', '8', '9', 'T', 'J', 'Q', 'K', 'A' }; char D[13] = { '2', '3', '4', '5', '6', '7', '8', '9', 'T', 'J', 'Q', 'K', 'A' }; char C[13] = { '2', '3', '4', '5', '6', '7', '8', '9', 'T', 'J', 'Q', 'K', 'A' }; /*4个玩家*/ char player1[13], player2[13], player3[13], player4[13]; int p1 = 0, p2 = 0, p3 = 0, p4 = 0; distribution(S, player1, player2, player3, player4, &p1, &p2, &p3, &p4); distribution(H, player1, player2, player3, player4, &p1, &p2, &p3, &p4); distribution(D, player1, player2, player3, player4, &p1, &p2, &p3, &p4); distribution(C, player1, player2, player3, player4, &p1, &p2, &p3, &p4); puts("运行结束"); for (int i = 0; i < 13; i++) printf("%c ", player1[i]); putchar('\n'); for (int i = 0; i < 13; i++) printf("%c ", player2[i]); putchar('\n'); for (int i = 0; i < 13; i++) printf("%c ", player3[i]); putchar('\n'); for (int i = 0; i < 13; i++) printf("%c ", player4[i]);}void distribution(char * S_H_D_C, char * player1, char * player2, char * player3, char * player4, int *p1, int *p2, int *p3, int *p4){ static int h = 1; int r; int a = *p1, b = *p2, c = *p3, d = *p4; for (int i = 0; i < 13; i++) { r = (rand() % 4) + 1; while ((r == 1 && (*p1) == 13) || (r == 2 && (*p2) == 13) || (r == 3 && (*p3) == 13) || (r == 4 && (*p4) == 13)) r = (rand() % 4) + 1; switch (r) { case 1: player1[(*p1)++] = S_H_D_C[i]; break; case 2: player2[(*p2)++] = S_H_D_C[i]; break; case 3: player3[(*p3)++] = S_H_D_C[i]; break; case 4: player4[(*p4)++] = S_H_D_C[i]; break; default: break; } } switch (h++) { case 1: printf("黑桃：\n"); break; case 2: printf("红桃：\n"); break; case 3: printf("方块：\n"); break; case 4: printf("梅花：\n"); break; } printf("Player1:"); for (int i = a; i < (*p1); i++) printf("%c ", player1[i]); putchar('\n'); printf("Player2:"); for (int i = b; i < (*p2); i++) printf("%c ", player2[i]); putchar('\n'); printf("Player3:"); for (int i = c; i < (*p3); i++) printf("%c ", player3[i]); putchar('\n'); printf("Player4:"); for (int i = d; i < (*p4); i++) printf("%c ", player4[i]); putchar('\n');} |
以下代码保证了当某个人得到13张牌后不在得牌
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r = (rand() % 4) + 1; while ((r == 1 && (*p1) == 13) || (r == 2 && (*p2) == 13) || (r == 3 && (*p3) == 13) || (r == 4 && (*p4) == 13)) r = (rand() % 4) + 1; |
以上就是本文的全部内容，希望对大家的学习有所帮助，也希望大家多多支持服务器之家。
原文链接：https://blog.csdn.net/weixin_43063069/article/details/90293134
