Dobry den,
toto je presne ono. Diky moc.
On Mon, 16 Apr 2007 11:08:01 +0200, Tomas Zverina <[EMAIL PROTECTED]>
wrote:
A neco jako:
Math.ceil(2*a) / 2
by nefungovalo?
Jan Medek napsal(a):
a = ( (a % 0.5) != 0 ) ? (Math.floor(a) + 0.5) : (a);
Ale nic moc. :-)
Petr Burdik napsal(a):
Ahojte,
lamu se ted s funci zaokrouhlovani na 50ti halere nahoru. Neresili
jste to prosim nekdo? Vim ze to hranici s cucanim, ale nejak se mi
nedari najit pekny algoritmus.
Pekny den
Pet
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