Code Golf: Playing Cubes

Answer 1

C, 287 characters

#define F(r)for(r=0;r<98;r++)
#define C(y,s)for(k=0;s[k];k++)b[49-i][w+k+y]=s[k];i++;
char k,i,j,w,t,b[98][99];main(c,v)char**v;{F(i)F(j)b[i][j]=32;
for(j=0;j0)t=i,i=0;for(;i

(this character count excludes two newlines)

This runs as a command line program, like

./cubes 1 2 3 4 5 0 1 3 2 -1 -10

You can try running it here: http://codepad.org/tu4HDqSy (This version is altered because codepad.org doesn't allow command line arguments.)

A handy tip for un-golfing is to run it through cpp and then indent:

char                k        , i, j, w, t, b[98][99];
main(c, v)
    char          **v;
{
    for (i = 0; i < 98; i++)
        for (j = 0; j < 98; j++)
            b[i][j] = 32;
    for (j = 0; j < c; j++) {
        i = 2 * atoi(v[j]);
        t = 0;
        if (i > 0)
            t = i, i = 0;
        for (; i < t; i -= 2) {
            for (k = 0; "|___|/"[k]; k++)
                b[49 - i][w + k + 0] = "|___|/"[k];
            i++;
            for (k = 0; "|   | |"[k]; k++)
                b[49 - i][w + k + 0] = "|   | |"[k];
            i++;
            for (k = 0; "/__ /|"[k]; k++)
                b[49 - i][w + k + 1] = "/__ /|"[k];
            i++;
            for (k = 0; "__"[k]; k++)
                b[49 - i][w + k + 3] = "__"[k];
            i++;
        } w += 4;
    } for (i = 0; i < 98; i++)
        puts(b[i]);
}