C语言自动发牌程序

C语言自动发牌程序

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：

1. 要设置数组表现扑克牌
2. 要设置数组表现玩家
3. 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：

1. 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
2. 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;