One fast option is to use BEDOPS:

$ bedops --intersect A.bed B.bed C.bed ... > answer.bed

You can use lots of inputs efficiently. The input files just have to be sorted.

The above command intersects B.bed C.bed etc. with A.bed, reporting all elements of A.bed that overlap B.bed C.bed etc.

Let's say you want to go the other direction efficiently. You can use BEDOPS with UNIX pipes and redirect standard output from one command to the next:

$ bedops --everything B.bed C.bed ... | bedops --intersect - A.bed > answer.bed

This does a multiset union of all the elements in B.bed C.bed etc. and passes these to an --intersect operation with A.bed.

The result file reports all elements of B.bed C.bed etc., which overlap A.bed.

The difference between these two directions is in which sets of elements get reported in the overlap. In the first case, elements of A.bed are reported. In the second case, elements of B.bed C.bed etc. are reported. Generally, this is not a symmetric operation.

If you have a lot of files to sort, a quick bash one-liner can take care of this:

$ for fn in `ls *.bed`; do sort-bed ${fn} > ${fn%.*}.sorted.bed; done

Some use GNU sort to do sorting of BED files, but BEDOPS sort-bed is usually faster.

latest bedtools version allows using wildcards, so finding the overlapping regions of all that 40 bed files with that particular 1 bed file would be as simple as:

bedtools intersect -a particular.bed -b all40*bed > all.40vs1.compared.bed

if you want to get the comparison for each one of the 40 with that particular 1, then you'll definitely have to loop:

for file in all40*bed; do
  bedtools intersect -a particular.bed -b $file > compared.$file
done