• 适合编程小白的C语言设计习题,实现自动发牌程序!源码分享!


    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('
    ');
    
    for (int i = 0; i < 13; i++)
    
    printf("%c ", player2[i]);
    
    putchar('
    ');
    
    for (int i = 0; i < 13; i++)
    
    printf("%c ", player3[i]);
    
    putchar('
    ');
    
    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:
    ");
    
      break;
    
    case 2:
    
      printf("红桃:
    ");
    
      break;
    
    case 3:
    
      printf("方块:
    ");
    
      break;
    
    case 4:
    
      printf("梅花:
    ");
    
      break;
    
    }
    
    printf("Player1:");
    
    for (int i = a; i < (*p1); i++)
    
      printf("%c ", player1[i]);
    
    putchar('
    ');
    
    printf("Player2:");
    
    for (int i = b; i < (*p2); i++)
    
      printf("%c ", player2[i]);
    
    putchar('
    ');
    
    printf("Player3:");
    
    for (int i = c; i < (*p3); i++)
    
      printf("%c ", player3[i]);
    
    putchar('
    ');
    
    printf("Player4:");
    
    for (int i = d; i < (*p4); i++)
    
      printf("%c ", player4[i]);
    
    putchar('
    ');
    
    }

    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我。


     

    不管你是转行也好,初学也罢,进阶也可——【值得关注】我的C/C++编程学习进阶俱乐部

    涉及到:C语言、C++、windows编程、网络编程、QT界面开发、Linux编程、游戏编程、黑客等等......

  • 相关阅读:
    .NET:CLR via C# A Brief Look at Metadata
    在容器中运行 Jenkins pipeline 任务
    Azure AI 服务之文本翻译
    linux journalctl 命令
    容器化的 DevOps 工作流
    System V IPC 之消息队列
    System V IPC 之信号量
    System V IPC 之共享内存
    减小容器镜像的三板斧
    linux chroot 命令
  • 原文地址:https://www.cnblogs.com/huya-edu/p/14052668.html
Copyright © 2020-2023  润新知