Question

我不确定如何让多维数组能够执行与我应该摆脱的其他三个相同的任务。我只是简单地用一个Multi替换所有其他数组的位置吗？这对我来说特别难，因为我不打牌，而纸牌游戏背后的逻辑对我来说毫无意义。

路线

我需要删除num_in_rank，num_in_suit和card_exists数组。让程序将卡存储在5 x 2阵列中。阵列的每一行都代表一张卡片。例如，如果数组名为hand，则hand（0）（0）将存储第一张牌的等级，而hand（0）（1）将存储第一张牌的套牌。

/* Classifies a poker hand */

#include <stdbool.h>   /* C99 only */
#include <stdio.h>
#include <stdlib.h>

#define NUM_RANKS 13
#define NUM_SUITS 4
#define NUM_CARDS 5

/* external variables */
int num_in_rank[NUM_RANKS];
int num_in_suit[NUM_SUITS];
bool straight, flush, four, three;
int pairs;   /* can be 0, 1, or 2 */
/* prototypes */
void read_cards(void);
void analyze_hand(void);
void print_result(void);

/**********************************************************
* main: Calls read_cards, analyze_hand, and print_result *
*       repeatedly.                                      *
**********************************************************/
int main(void)
{
for (;;) {
    read_cards();
    analyze_hand();
    print_result();
}
}
/**********************************************************
* read_cards: Reads the cards into the external          *
*             variables num_in_rank and num_in_suit;     *
*             checks for bad cards and duplicate cards.  *
**********************************************************/
void read_cards(void)
{
bool card_exists[NUM_RANKS][NUM_SUITS];
char ch, rank_ch, suit_ch;
int rank, suit;
bool bad_card;
int cards_read = 0;

for (rank = 0; rank < NUM_RANKS; rank++) {
    num_in_rank[rank] = 0;
    for (suit = 0; suit < NUM_SUITS; suit++)
        card_exists[rank][suit] = false;
}

for (suit = 0; suit < NUM_SUITS; suit++)
    num_in_suit[suit] = 0;
while (cards_read < NUM_CARDS) {
    bad_card = false;

    printf("Enter a card: ");

    rank_ch = getchar();
    switch (rank_ch) {
    case '0':           exit(EXIT_SUCCESS);
    case '2':           rank = 0; break;
    case '3':           rank = 1; break;
    case '4':           rank = 2; break;
    case '5':           rank = 3; break;
    case '6':           rank = 4; break;
    case '7':           rank = 5; break;
    case '8':           rank = 6; break;
    case '9':           rank = 7; break;
    case 't': case 'T': rank = 8; break;
    case 'j': case 'J': rank = 9; break;
    case 'q': case 'Q': rank = 10; break;
    case 'k': case 'K': rank = 11; break;
    case 'a': case 'A': rank = 12; break;
    default:            bad_card = true;
    }
    suit_ch = getchar();
    switch (suit_ch) {
    case 'c': case 'C': suit = 0; break;
    case 'd': case 'D': suit = 1; break;
    case 'h': case 'H': suit = 2; break;
    case 's': case 'S': suit = 3; break;
    default:            bad_card = true;
    }

    while ((ch = getchar()) != '\n')
    if (ch != ' ') bad_card = true;

    if (bad_card)
        printf("Bad card; ignored.\n");
    else if (card_exists[rank][suit])
        printf("Duplicate card; ignored.\n");
    else {
        num_in_rank[rank]++;
        num_in_suit[suit]++;
        card_exists[rank][suit] = true;
        cards_read++;
    }
}
}
/**********************************************************
* analyze_hand: Determines whether the hand contains a   *
*               straight, a flush, four-of-a-kind,       *
*               and/or three-of-a-kind; determines the   *
*               number of pairs; stores the results into *
*               the external variables straight, flush,  *
*               four, three, and pairs.                  *
**********************************************************/
void analyze_hand(void)
{
int num_consec = 0;
int rank, suit;
straight = false;
flush = false;
four = false;
three = false;
pairs = 0;
/* check for flush */
for (suit = 0; suit < NUM_SUITS; suit++)
if (num_in_suit[suit] == NUM_CARDS)
    flush = true;

/* check for straight */
rank = 0;
while (num_in_rank[rank] == 0) rank++;
for (; rank < NUM_RANKS && num_in_rank[rank] > 0; rank++)
    num_consec++;
if (num_consec == NUM_CARDS) {
    straight = true;
    return;
}

/* check for 4-of-a-kind, 3-of-a-kind, and pairs */
for (rank = 0; rank < NUM_RANKS; rank++) {
    if (num_in_rank[rank] == 4) four = true;
    if (num_in_rank[rank] == 3) three = true;
    if (num_in_rank[rank] == 2) pairs++;
}
}
/**********************************************************
* print_result: Prints the classification of the hand,   *
*               based on the values of the external      *
*               variables straight, flush, four, three,  *
*               and pairs.                               *
**********************************************************/
void print_result(void)
{
if (straight && flush) printf("Straight flush");
else if (four)         printf("Four of a kind");
else if (three &&
    pairs == 1)   printf("Full house");
else if (flush)        printf("Flush");
else if (straight)     printf("Straight");
else if (three)        printf("Three of a kind");
else if (pairs == 2)   printf("Two pairs");
else if (pairs == 1)   printf("Pair");
else                   printf("High card");

printf("\n\n");
}

Answer 1

帮助您可视化它首先定义多维数组，然后使用常量来访问索引

即

//you could also use namespaces and enums or static class variables, const is simpler
const int RANK = 0; //or CARD_RANK if ambiguous/conflicting
const int SUIT = 1; //or CARD_SUIT if ambiguous
int cards[5][2]; //first index is card, second index is data about the card

//access first card's suit
int temp = cards[0][SUIT];

假设程序已经运行，你需要做的就是这个，那就去做吧，一切都会好起来的。这场比赛似乎是扑克或变种btw。

扑克计划中的多维数组

1 个答案: