適合程式設計小白的C語言設計習題，實現自動發牌程式！原始碼分享！

阿新 • • 發佈：2020-11-28

C語言自動發牌程式，供大家參考，具體內容如下：

一副撲克有52張牌，打橋牌時應將牌分給4個人。請設計一個程式完成自動發牌的工作。要求：黑桃用S (Spaces)表示，紅桃用H (Hearts)表示，方塊用D (Diamonds)表示，梅花用C (Clubs)表示。

分析：

要設定陣列表現撲克牌

要設定陣列表現玩家

要給撲克牌做特定標識，得到結果後玩家要知道自己手中黑桃有哪些、方塊有哪些

初步想法：

設定4個字元陣列儲存4種梅花牌，設定4個字元陣列表示4名玩家分配到的牌

每張牌隨機發給4名玩家，當玩家的持牌數達到13，不再分配給該名玩家牌

程式碼展示：

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個玩家*/

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("執行結束");

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("紅桃：\n");

  break;

case 3:

  printf("方塊：\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');

}

以下程式碼保證了當某個人得到13張牌後不在得牌：

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;