top 명령어에서 cpu 점유율 구하는 방법 혹시 아시나요?

#!/usr/bin/perl
 
use strict;
 
my $__used   = 1;
my $__nice   = 2;
my $__sys    = 3;
my $__idle   = 4;
my $__iowait = 5;
my $__cxxx   = 6;
my $__cyyy   = 7;
 
my $_user = 0;
my $_sys  = 1;
my $_idle = 2;
my $_iowa = 3;
my $_ext  = 4;
 
my @cpus_prev;
my @cpus_post;
 
@cpus_prev = getstat ();
 
sleep (2);
 
@cpus_post = getstat ();
 
my $user = $cpus_post[$_user] - $cpus_prev[$_user];
my $sys  = $cpus_post[$_sys] - $cpus_prev[$_sys];
my $idle = $cpus_post[$_idle] - $cpus_prev[$_idle];
my $iowa = $cpus_post[$_iowa] - $cpus_prev[$_iowa];
my $ext  = $cpus_post[$_ext]  - $cpus_prev[$_ext];
 
my $div = $user + $sys + $idle + $iowa;
my $scale = 100.0 / $div;
 
printf ("%.2f : %.2f : %.2f\n", fixed($user), fixed ($sys), fixed ($idle));
 
 
sub fixed {
    my ($i) = @_;
    return ($i * $scale);
}
 
sub getstat {
    my @_cpu;
    my @_proc;
 
    my $__proc = `cat /proc/stat | grep cpu0`;
    chomp ($__proc);
    @_proc = split /[\s]+/, $__proc;
 
    # for 2.4 kernel
    $_proc[$__iowait] = 0;
    $_proc[$__cxxx] = 0;
    $_proc[$__cyyy] = 0;
 
    $_cpu[$_user] = $_proc[$__used] + $_proc[$__nice];
    $_cpu[$_sys]  = $_proc[$__sys] + $_proc[$__cxxx] + $_proc[$__cyyy];
    $_cpu[$_idle] = $_proc[$__idle];
    $_cpu[$_iowa] = $_proc[$__iowait];
 
    return @_cpu;
}
 
 
exit (0);