First of all nobody should ever use csh for anything - it's outdated and un- (not "under") powered. Secondly, I doubt it's up to the task. Third, it's much more likely that awk, sed or even Perl will be a much better tool for the task.
awk '/^#/ {printf line; line=$0"\n"; next} /pattern/ {line=""} ! /pattern/ {printf line; print; line=""}'
Edit: fixed script to handle comment lines correctly
sed -n '10000000,10000020p' filename
You might be able to speed that up a little like this:
sed -n '10000000,10000020p; 10000021q' filename
In those commands, the option -n causes sed to "suppress automatic printing of pattern space". The p command "print[s] the current pattern space" and the q command "Immediately quit[s] the sed script without processing any more input..." The quotes are from the sed man page.
By the way, your command
tail -n 10000000 filename | head 10
starts at the ten millionth line from the end of the file, while your "middle" command would seem to start at the ten millionth from the beginning which would be equivalent to:
head -n 10000010 filename | tail 10
The problem is that for unsorted files with variable length lines any process is going to have to go through the file counting newlines. There's no way to shortcut that.
If, however, the file is sorted (a log file with timestamps, for example) or has fixed length lines, then you can seek into the file based on a byte position. In the log file example, you could do a binary search for a range of times as my Python script here* does. In the case of the fixed record length file, it's really easy. You just seek linelength * linecount characters into the file.
* I keep meaning to post yet another update to that script. Maybe I'll get around to it one of these days.
Best Answer
TO show how many lines containing foo are there.