如何在C中查找，計數和顯示卡對，其中2個或3個是一對，而4個是2對呢？

Question

我在掙扎，不知道如何確定和顯示對數

• 種類2或3是一對，種類4是2對。
• 種類2是2個具有相同編號的卡，種類3是3具有相同編號的卡，種類4是4具有相同編號的卡。

由於卡片組中有52張卡片，您將如何查找，計算和顯示對數？

碼：

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

/* handy typedefs */
typedef unsigned char card;
typedef unsigned char pairs;

/* arrays for the names of things */
static char *suits[4] = {"Hearts","Diamonds","Clubs","Spades"};
static char *values[13]= {"Ace","Two","Three","Four","Five","Six","Seven",
                      "Eight","Nine","Ten","Jack","Queen","King"};
static char *colour[2]= {"Red","Black"};

int compareface(const void * c1, const void * c2);
int comparesuit(const void * c1, const void * c2);
void shuffle(int deck[52]);
pairs findpairs ( card *h ) ; /* finds any pairs in a hand */
int main()
{
 int deck[52];
 pairs numpairs[5], highest;
 int s, c, a, i, j;

 j = 0;
 for(s = 0; s < 4; s++)//for filling a deck of 52 cards
 {
   for(c = 0; c < 13; c++)
   {
   deck[s * 13 + c] = j;
   j++;
   }
 }

for(a = 0; a < 52; a++)
{     
  printf(" %s of", values[ deck[a] % 13 ]);
  printf(" %s, is ", suits[ deck[a] / 13 ]);
  printf("%s\n", colour[ deck[a] / 26 ]);
}

int hands[5][5],h,cd,winner,handssorted[5][5];
int irand;
srand(time(NULL));       /* seed the random number generator */

// shuffle the deck before to get the hands:
shuffle(deck);
j = 0;
for (h=0;h<5;h++)
{
  for ( i = 0; i < 5; i++) {
    hands[h][i] = deck[j];
    j++;
  }
}
// sort the cards by card value:
for (h=0;h<5;h++)
{
  qsort(hands[h], 5, sizeof(int), compareface);

}
// print the hand:
for (h=0;h<5;h++)
{
  printf("\nHand %d:\n",h+1 );
  for ( i = 0; i < 5; i++) {
    printf(" %s of", values[ hands[h][i] % 13 ]);
    printf(" %s, is ", suits[ hands[h][i] / 13 ]);
    printf("%s\n", colour[ hands[h][i] / 26 ]);
  }
 printf("Number of pairs: \n");
}

// sort the cards by card value then card suit:
for (h=0;h<5;h++)
{
  qsort(hands[h], 5, sizeof(int), comparesuit);
  numpairs[h]= findpairs( handssorted[h] );
}
// print the hand:
  for (h=0;h<5;h++)
  {
   for ( i = 0; i < 4; i++) {
    for(j = i+1; j < 5; j++) {
     if( (hands[h][i] / 13 > hands[h][j] / 13) && (hands[h][i] % 13 ==  hands[h][j] % 13) ) {
      c = hands[h][i];
      hands[h][i] = hands[h][j];
      hands[h][j] = c;
                                         }
                         }
                       }
              }
       /* determine the winner and print it */
       return 0;
        }

     void shuffle(int deck[52])
     {
      int i,rnd, c;

      for(i=0;i<52;i++)
      {
        /* generate a random number between 0 & 51 */
         rnd=rand() * 52.0 / RAND_MAX;
         c = deck[i];
         deck[i] = deck[rnd];
         deck[rnd] = c;
        }
      }

    int compareface(const void * c1, const void * c2)
    {
    const int cd1 = *(const int*)c1;
    const int cd2 = *(const int*)c2;
    if(cd1 % 13 > cd2 % 13) return 1;
    if(cd1 % 13 == cd2 % 13) return 0;
     return -1;
    }

    int comparesuit(const void * c1, const void * c2)
     {
     const int cd1 = *(const int*)c1;
     const int cd2 = *(const int*)c2;
     if(cd1 / 13 > cd2 / 13) return 1;
     if(cd1 / 13 == cd2 / 13) return 0;
     return -1;
     }

      pairs findpairs ( card *h )
      {
       pairs numpairs=0;
       /* find the pairs here */
       return numpairs;
        }

Answer 1

卡片排序后，只需比較連續的卡片就可以找到對。 在此之前，您的代碼中還有另一個問題，當您進行排序時，您是在hands [h]中對其進行排序，而對於findpairs（）來說，您是在給手排序數組。 因此，請先將這些值復制到手動排序后再進行排序。

之后，您只需比較findpairs調用中的連續值即可。 找到對后，將指針增加到第三張卡。

 pairs findpairs ( card *h )
  {
   pairs numpairs=0;
   int card1, card2;
   /* find the pairs here */
    for(int i = 0; i<4; i++){
       /* Compare the card values here */
        card1 = h[i];
        card2 = h[i++]; /*extract values according to your encoding scheme*/
       if(card1 == card2){
          numpairs++;
          i++; // skip the next card to avoid counting 3 of a kind, since it is just a pair 
     }

   return numpairs;
    }