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！
2021-03-13 16:32
标签：nbsp 适合 height sign pre data rgba 字符数组 标识 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。 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 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 以下代码保证了当某个人得到13张牌后不在得牌： 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...... 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！ 标签：nbsp 适合 height sign pre data rgba 字符数组 标识 原文地址：https://www.cnblogs.com/huya-edu/p/14052668.html分析：
初步想法：
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 )
printf("%c ", player1[i]);
putchar(‘\n‘);
printf("Player2:");
for (int i = b; i )
printf("%c ", player2[i]);
putchar(‘\n‘);
printf("Player3:");
for (int i = c; i )
printf("%c ", player3[i]);
putchar(‘\n‘);
printf("Player4:");
for (int i = d; i )
printf("%c ", player4[i]);
putchar(‘\n‘);
}
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;
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文章标题：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！
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