LaTeX makro: 1 bod, 2 body, 5 bodu?

[Michal Kubecek]

> > Nebo elegantneji, abyste se obesel bez slozenych zavorek (aby
> > stacilo \b2, \b18):
> >
> > \def\b{\bgroup\afterassignment\bb\count0=}
> > \def\bb{%
> >   \the\count0\ bod%
>     \count1=\count0
>     \divide\count1 by 10
>     \multiply\count1 by 10
>     \advance\count0 by -\count1
> >   \ifnum\count0>4 u\else\ifnum\count0>1 y\fi\fi
> >   \egroup
> > }

Myslim, ze tohle bude davat spatny vysledek pro cisla koncici
na 11, 12, 13, 14. Prave proto jsem se tomu chtel vyhnout.
Takze pro totalni zmateni jeste

> > \def\b{\bgroup\afterassignment\bb\count0=}
> > \def\bb{%
> >   \the\count0\ bod%
>     \count1=\count0
>     \divide\count1 by 10
>     \multiply\count1 by 10
      \count2=\count0
>     \advance\count0 by -\count1
      \count1=\count2
      \divide\count1 by 10
      \multiply\count1 by 10
      \advance\count2 by -\count1
      \ifnum\count1=1 u\else
> >     \ifnum\count0>4 u\else\ifnum\count0>1 y\fi\fi
      \fi
> >   \egroup
> > }


> Znak = a klicova slova by jsou nepovinna.

Krome toho '=' na prvnim radku.

                                                Michal Kubecek