How to print lines till till a pattern is matched in loop

Shell Programming and Scripting
1

Dear All

I have a file like this

112534554
446538656
444695656
225696966
226569744
228787874
113536566
443533535
222564552
115464656
225445345
225533234

I want to cut the file into different parts where the first two columns are '11' . The first two columns will be either 11, 22,44 or 66.

I want the output to be directed to three files named serially like f001, f002, f003.......like this

cat f001
112534554
446538656
444695656
225696966
226569744
228787874

cat f002
113536566
443533535
222564552

cat f003
115464656
225445345
225533234

How can i put this in a loop till the end of read of my input file......

Please help.....

Regards

2 
awk '/^11/{++c}{print > "f00" c}' file

In case there are too many split files use,

awk '/^11/{++c;close(f)}{print > (f="f00" c)}' file
3

Thanks....!!!
The second comand is working perfectly....

4

awk:

nawk '/11.*/{n++;f=sprintf("a%s.txt",n);print $0 > f;next}{print $0
>f}' file

perl:

undef $/;
open FH,"<file" or die "Can not open";
$str=<FH>;
$str=~s/\n11/,11/g;
@arr=split(",",$str);
close FH;
for($i=0;$i<=$#arr;$i++){
	$file=sprintf("f%s.txt",$i+1);
	open FH,">$file";
	print FH $arr[$i];
	close FH;
}