Perl : Improve this group of if statements ?=

[plenderj]

I have the following piece of code in perl.
I'm sure there must be a better way of doing this... but I don't know very much about perl.

Any ideas ?

if ($ipAddress[0] == 61)  {
	$which = 2;
} elsif ($ipAddress[0] == 62) {
	$which = 0;
} elsif (($ipAddress[0] >= 63) && ($ipAddress[0] <= 69))   {
	$which = 1;
} elsif (($ipAddress[0] == 80) || ($ipAddress[0] == 81))   {
	$which = 0;
} elsif (($ipAddress[0] >= 128) && ($ipAddress[0] <= 192)) {
	$which = 1;
} elsif (($ipAddress[0] >= 193) && ($ipAddress[0] <= 195)) {
	$which = 0;
} elsif ($ipAddress[0] == 196) {
	$which = 3;
} elsif ($ipAddress[0] == 198) {
	$which = 1;
} elsif (($ipAddress[0] >= 193) && ($ipAddress[0] <= 195)) {
	$which = 0;
} elsif (($ipAddress[0] >= 199) && ($ipAddress[0] <= 201)) {
	$which = 1;
} elsif (($ipAddress[0] >= 202) && ($ipAddress[0] <= 203)) {
	$which = 2;
} elsif (($ipAddress[0] >= 204) && ($ipAddress[0] <= 209)) {
	$which = 1;
} elsif (($ipAddress[0] >= 210) && ($ipAddress[0] <= 211)) {
	$which = 2;
} elsif (($ipAddress[0] >= 212) && ($ipAddress[0] <= 213)) {
	$which = 0;
} elsif (($ipAddress[0] >= 214) && ($ipAddress[0] <= 215)) {
	$which = 1;
} elsif ($ipAddress[0] == 216) {
	$which = 1;
} elsif ($ipAddress[0] == 217) {
	$which = 0;
} elsif (($ipAddress[0] >= 218) && ($ipAddress[0] <= 221)) {
	$which = 2;
} elsif ($ipAddress >= 222) {
	$which = 1;
}

[john tindell]

i dont know perl, but i think its kinda like C

so could you try this.

switch($ipAddress[0])
{
case 62
$which = 0;
break;
}

dont hold me to this though