中国将帅象棋问题

#include <stdio.h>

#define HALF_BITS_LENGTH 4
#define FULLMASK 255    // 1111 1111
#define LMASK (FULLMASK << HALF_BITS_LENGTH)    // 1111 0000
#define RMASK (FULLMASK >> HALF_BITS_LENGTH) //0000 1111
// 低四位置为 n 的低四位
#define RSET(b, n) (b = ((LMASK & b) | (n)))
// 高四位置为 n 的高四位
#define LSET(b, n) (b = ((RMASK & b) | ( n << HALF_BITS_LENGTH)))
// 获得 b 的低四位,输出时候为8位的char,高四位为0
#define RGET(b) (RMASK & b)
// 获得 b 的高四位, 输出时候为8位的char,高四位为0
#define LGET(b) ((LMASK & b) >> HALF_BITS_LENGTH)
#define GRIDW 3

int main() {
        unsigned char b;
        for(LSET(b, 1); LGET(b) <= GRIDW * GRIDW; LSET(b, (LGET(b) + 1))) {
                for(RSET(b, 1) ; RGET(b) <= GRIDW * GRIDW; RSET(b, (RGET(b) + 1))) {
                        if(LGET(b) % GRIDW != RGET(b) % GRIDW)
                                printf("A = %d, B = %d\n", LGET(b), RGET(b));
                }
        }
        return 0;
}

A = 1, B = 2
A = 1, B = 3
A = 1, B = 5
A = 1, B = 6
A = 1, B = 8
A = 1, B = 9
A = 2, B = 1
A = 2, B = 3
A = 2, B = 4
A = 2, B = 6
A = 2, B = 7
A = 2, B = 9
A = 3, B = 1
A = 3, B = 2
A = 3, B = 4
A = 3, B = 5
A = 3, B = 7
A = 3, B = 8
A = 4, B = 2
A = 4, B = 3
A = 4, B = 5
A = 4, B = 6
A = 4, B = 8
A = 4, B = 9
A = 5, B = 1
A = 5, B = 3
A = 5, B = 4
A = 5, B = 6
A = 5, B = 7
A = 5, B = 9
A = 6, B = 1
A = 6, B = 2
A = 6, B = 4
A = 6, B = 5
A = 6, B = 7
A = 6, B = 8
A = 7, B = 2
A = 7, B = 3
A = 7, B = 5
A = 7, B = 6
A = 7, B = 8
A = 7, B = 9
A = 8, B = 1
A = 8, B = 3
A = 8, B = 4
A = 8, B = 6
A = 8, B = 7
A = 8, B = 9
A = 9, B = 1
A = 9, B = 2
A = 9, B = 4
A = 9, B = 5
A = 9, B = 7
A = 9, B = 8