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我在我的计算机科学课上有一项作业(我几周前发布了这个问题,但它的解释方式不适合我应该做的程序)。我已经有一个程序可以在这里洗牌和发牌。

#include <stdio.h>
#include <stdlib.h>
#include <time.h>

void shuffle( int [][ 13 ] );
void deal ( const int[][ 13 ], const char *[], const char *[] );


 int main()
 {
  const char *suit[4] ={"Hearts","Diamonds","Clubs","Spades"};
    const char *face[13] ={"Ace", "Duece", "Three", "Four", "Five", "Six", 
    "Seven", "Eight","Nine", "Ten", "Jack", "Queen", "King"};
    int deck[4][13] = {0};
    int row, column, card = 1;


    for( row = 0; row <= 3; row++ )
    {
        for(column = 0; column <= 12; column++)
            {
            deck[row][column] = card;
            card++;
            }
    }

        srand(time(0));
        shuffle(deck);
        deal(deck, face, suit);
    return 0;
 }

 void shuffle( int wDeck[][13] )
 {
    int row, column, randomColumn, randomRow, card = 1, counter1, counter2, hold;




    for( counter1 = 0; counter1 <= 3; counter1++) 
    {
        for(counter2 = 0; counter2 <= 12; counter2++)
            {
                randomColumn = rand() % 13;
                randomRow = rand() % 4;

                    hold = wDeck[counter1][counter2];
                    wDeck[counter1][counter2] = wDeck[randomRow][randomColumn]; 
                    wDeck[randomRow][randomColumn] = hold;
            }
       }
    }

 void deal( const int wDeck[][13], const char *wFace[], const char *wSuit[] )
{
  int card, row, column;

        for ( card = 1; card <= 52; card++ )
            for (row = 0; row <= 3; row++ )
                for ( column = 0; column <= 12; column++ )
                    if( wDeck[row][column] == card )
                    {
                    printf("%5s of %-8s%c",wFace[ column ], wSuit[row], card % 2 == 0 ? '\n' : '\t');
                    break;
                    }
}

我应该修改发牌功能以发一张 5 张牌的扑克手,然后检查他们有什么“等级”扑克手(两种,同花)。我的老师提到创建一个单独的双脚本数组来做到这一点,但我可以用不同的方式来做。问题是,我必须使用当前的牌组/洗牌设置来做到这一点。谁能解释如何做到这一点?如果它效率低下也没关系,只要它有效。

4

1 回答 1

1

对于现有的deal原型,没有办法从它返回信息——它没有返回值,所有的参数都是const. 由于牌组已经洗过(由shuffle()),我假设这个想法是让发牌函数查看牌组数组中的前五张牌,并找到它们的等级。这可以完全在deal函数内完成,或者(更好)通过int rank(int hand[5])从函数内调用函数来完成deal

于 2013-04-20T08:30:05.280 回答