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，供你们参考，具体内容以下：

一副扑克有52张牌，打桥牌时应将牌分给4我的。请设计一个程序完成自动发牌的工做。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

要给扑克牌作特定标识，获得结果后玩家要知道本身手中黑桃有哪些、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，再也不分配给该名玩家牌

代码展现：

void mycode_13()

{

srand(unsigned(time(NULL)));

/*所有牌*/

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');

}

 

如下代码保证了当某我的获得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;

以上就是本文的所有内容，但愿对你们的学习有所帮助，也但愿你们多多支持我。

 

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