C语言实现自动发牌程序
作者：陈子剑
这篇文章主要为大家详细介绍了C语言实现自动发牌程序，文中示例代码介绍的非常详细，具有一定的参考价值，感兴趣的小伙伴们可以参考一下
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脚本之家。