Problem 1：1000一下，能被3，5整除的数的和

If we list all the natural numbers below 10 that are multiples of 3 or 5,we get 3,5,6 and 9. The sum of these multiples is 23.

Find the sum of all the multiples of 3 or 5 below 1000.

my $sum=0;
foreach(1..999)
{
	$sum=$sum+$_ if(0==$_%3||0==$_%5);
}
print $sum."n";

Problem 2，Fibonacci数列

use strict;
use warnings;

my $first_value =1;
my $second_value=2;
my $third_value =0;
my $sum         =2;

while($third_value<4000000)
{
	$third_value=$first_value+$second_value;
	if(0==$third_value%2)
	{
		$sum=$sum+$third_value;
	}
	$first_value =$second_value;
	$second_value=$third_value ;
}
print $sum;