perl将时间的09取成9

[theone5288]

问题1--我是菜鸟，现在有这样一个数据，2 03：20：01，我想依次取出2，3，20，1；其中凡是01，02等都取出1，2，去掉前面的0！
问题2--其实数据2 03：20：01是个时间，即2天3小时20分1秒，有没有办法直接换算成分钟？能直接换算最好，不行的话就用上面的方法依次取出数做相应的运算！
求教~~~~~~~~~~~

[iambic]

my $time = "2 03:20:01";
my ($day, $hour, $minute, $second) = ($time =~ /\d+/g);
my $minutes = ($day * 24 + $hour) * 60 + $minute;
print $minutes;

[theone5288]

$time = "2 03:20:01";
my @time = split(' ',$time);
my $day = $time[0]*24*60;
my @time1 = split(':',$time[1]);
my $i;
for ($i=0;$i<=$#time1;$i++)
{
$time1[$i] =~ s/^0//;
print $time1[$i],"\n";

}
my $hour = $time1[0]*60;
my $minute = $time1[1];
my $sum_minute = $day + $hour + $minute;
print $sum_minute,"\n";
$sum_minute 就是所换算玩的分钟数

[njlnb2000]

"2 03:20:01"=~/(\d+) +(\d+):(\d+)/;
print (($1*24+$2)*60+$3);

[njlnb2000]

#如果非要去0的话可以这样：
"2 09:20:01"=~/(\d+) +0?(\d+):0?(\d+):0?(\d+)/;
print (($1*24+$2)*60+$3,"\n");#分钟数
print "$1 $2:$3:$4\n";#去0之后按格式输出

[proorck6]

$time<10?"0".$time:$time;

[lixq2000]

"%02d"%(3)
'03'