#!/usr/bin/perl
$n1 = 1;
$n2 = 7;
$zing=5;
$inter=(($n2-$n1+1)/$zing);
for ($i=$n1;$i<=$n2;$i=$i+$inter) {
$pirm=int($i+0.5);
$sek=int($i+$inter+0.5)-1;
if ($sek > $n2) {$sek=$n2;}
print "$pirm - $sek\n";
}

1 - 1
2 - 3
4 - 4
5 - 6
7 - 7
taip netinka
"Bamba" <none@nonde.none> wrote in message 
news:go5nhv$6og$1@trimpas.omnitel.net...
> matai ner taip paprasta kaip atrodo :)
>
> grupiu negali but daugiau nei $zing (5).
>
> yra dar variantai kaip 1-7
>
> tai turetu pasiskirtyti taip:
>
> 1-2
> 3-4
> 5-5
> 6-6
> 7-7
>
> "Nerijus T" <news@upe.lt> wrote in message 
> news:go5mqs$5b3$1@trimpas.omnitel.net...
>> #!/usr/bin/perl
>> $n1 = 1;
>> $n2 = 19;
>> $zing=5;
>> print $a;
>> $liek=($n2-$n1+1)%$zing;
>> $p=0;
>> if ($liek<=>0) {$p=1};
>> $inter=int(($n2-$n1+1)/($zing-$p));
>> for ($i=$n1;$i<=$n2;$i=$i+$inter) {
>> $sek=$i+$inter-1;
>> if ($sek > $n2) {$sek=$n2;}
>> print "$i - $sek\n";
>> }
>>
>>
>> "Nerijus T" <news@upe.lt> wrote in message 
>> news:go5l8n$29h$1@trimpas.omnitel.net...
>>> zingsniui paskaiciuoti galima naudoti sveiko skaiciaus be apvalinimo 
>>> iskirimo funkcija int
>>> toliau mastyk pats
>>> "Bamba" <none@nonde.none> wrote in message 
>>> news:go5imt$t3h$1@trimpas.omnitel.net...
>>>> Sveiki,
>>>>
>>>> situacija tokia:
>>>>
>>>> yra diapazonas sakykim nuo 1-19.
>>>>
>>>> Reikia kokio neskausmingo budo sita diapazonas padalinti i pvz 5 
>>>> grupes.
>>>>
>>>>
>>>> 1-4
>>>> 5-8
>>>> 9-12
>>>> 13-16
>>>> 17-19
>>>>
>>>>
>>>> pvz 1-9
>>>>
>>>> 1-2
>>>> 3-4
>>>> 5-6
>>>> 7-8
>>>> 9-9
>>>>
>>>> Gal kas turit kokiu ideju?
>>>>
>>>
>>>
>>
>>
>
>