counting sort + bit operation

UVa problem: Bridge Hands

每一花色牌不超过13张，4种花色，用一个long long表示，每16位一组表示一个花色

#include <stdio.h>
#include <string.h>

long long player[4];
int p[90];
char ip[4] = {'C', 'D', 'S', 'H'};
int seat[90];
char iseat[4] = {'S','W','N','E'};
int idx[90];
char inverse_idx[15] = {'0','1','2','3','4','5','6','7','8','9','T','J','Q','K','A'};
//int __builtin_ctz(long long);
int main() {
    char buff[60];
    seat['S'] = p['C'] = 0; 
    seat['W'] = p['D'] = 1;
    seat['N'] = p['S'] = 2;
    seat['E'] = p['H'] = 3;
    int i, j;
    for(i = 2; i <= 9; ++i)
        idx[i+48] = i;
    idx['T'] = 10;
    idx['J'] = 11;
    idx['Q'] = 12;
    idx['K'] = 13;
    idx['A'] = 14;       
    char t;
    while((t = getchar()) != '#') {
        while(gets(buff) && strlen(buff)==0){};
        int side = seat[t];
        memset(player, 0, sizeof(player));
        for(i = 0; i < 52; i += 2) {
            if(++side == 4) side = 0;
            /* get card, set its bit */
            player[side] |= 1LL<< (p[buff[i]]<<4LL) + idx[buff[i+1]]; 
        }
        gets(buff);
        for(i = 0; i < 52; i += 2) {
            if(++side == 4) side = 0;
            player[side] |= 1LL<< (p[buff[i]]<<4LL) + idx[buff[i+1]];
        }
        
        for(i = 0; i < 4; ++i) {
            int printed = 0;
            printf("%c: ", iseat[i]); 
            for(j = 0; j < 4; ++j) {                
                /* get current color */
                int c = player[i] & 0xffff;
                player[i] >>= 16; /* next color */
                while(c) {
                    int pos = __builtin_ctz(c);
                    c ^= 1LL << pos;  /* clear printed bit */
                    printf("%c%c", ip[j], inverse_idx[pos]); 
                    if(++printed < 13) printf(" ");
                }
            }
            printf("\n");
        }
    }
}

/* gcc has built in __builtin_ctz, vc not*/ 

/*
int __builtin_ctz(long long c) {

    for(int i = 0; i < 64; ++i)
        if(c & (1LL<<i)) return i;
}*/