C语言实现自动发牌程序

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我是靠谱客的博主灵巧故事，最近开发中收集的这篇文章主要介绍C语言实现自动发牌程序，觉得挺不错的，现在分享给大家，希望可以做个参考。

概述

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：

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

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靠谱客。