一个用 C 语言写的迷你版 2048 游戏，仅仅有 500个字符

Jay Chan 用 C 语言写的一个迷你版 2048 游戏，仅仅有 487 个字符。

来围观吧

M[16],X=16,W,k;main(){T(system("stty cbreak")
);puts(W&1?
"WIN":"LOSE");}K[]={2,3,1};s(f,d,i
,j,l,P){for(i=4;i--;)for(j=k=l=0;k<4;)j<4?
P=M
[w(d,i,j++)],W|=P>>11,l*P&&(f?M[w(d,i,k)]=l<<
(l==P):0,k++),l=l?P?l-P?P:0:l:P:(f?M[w(d,i,k)
]=l:0,++k,W|=2*!l,l=0);}w(d,i,j){return d?w(d
-1,j,3-i):4*i+j;}T(i){for(i=X+rand()%X;M[i%X]
*i;i--);i?
M[i%X]=2<<rand()%2:0;for(W=i=0;i<4;
)s(0,i++);for(i=X,puts("e[2Je[H");i--;i%4||
puts(""))printf(M[i]?"%4d|":"    |",M[i]);W-2
||read(0,&k,3)|T(s (1,K[(k>>X)%4]));}

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

#define GRID_LEN 16

int M[GRID_LEN];
int X = GRID_LEN;
int W;
int k;
int K[] = { 2, 3, 1 };


int w (int d, int i, int j)
{
  if (d <= 0) {
    return 4 * i + j;
  }

  return w (d - 1, j, 3 - i);
}

void s (int f, int d)
{
  int i = 4, j, l, P;

  for (; i--;) {
    j = k = l = 0;

    for (; k < 4;) {
      if (j < 4) {
        P = M[w (d, i, j++)];
        W |= P >> 11;
        l *P && (f ? M[w (d, i, k)] = l << (l == P) : 0, k++);
        l = l ? (P ? (l - P ?
 P : 0) : l) : P;
      }
      else {
        f ? M[w (d, i, k)] = l : 0;
        ++k;
        W |= 2 * !l;
        l = 0;
      }
    }
  }
}

void T ()
{
  int i = X + rand () % X;

  for (; M[i % X] * i; i--);

  i ? M[i % X] = 2 << rand () % 2 : 0;
  W = i = 0;

  for (; i < 4; i++) {
    s (0, i);
  }

  // Prints the tiles onto the terminal
  i = X;
  puts ("e[2Je[H");

  for (; i--;) {
    if (M[i]) {
      printf ("%4d|", M[i]);
    } else {
      printf ("%s", "    |");
    }

    // every 4th cell is followed by a line-break
    if (0 == (i & 3)) {
      putchar ('
');
    }
  }

  // read input from keyboard
  if (!(W - 2)) {
    read (0, &k, 3);
    s (1, K[(k >> X) % 4]);
    T ();
  }
}

int main (void)
{
  // Uses stty to clear the screen in preparation for the game
  //system ("stty cbreak");

  /* Intializes random number generator */
  srand ((unsigned) time (NULL));

  T ();

  // Game has finished by this point
  // If win, display "WIN". Otherwise, display "LOSE".
  puts (W & 1 ? "WIN" : "LOSE");

  return 0;
}