Aceguy Posted January 6, 2008 Share Posted January 6, 2008 please help. ( a tough one i know) [u]My Projects.[/u]Launcher - not just for games & Apps (Mp3's & Network Files)Mp3 File RenamerMy File Backup UtilityFFXI - Realtime to Vana time Clock Link to comment Share on other sites More sharing options...
PsaltyDS Posted January 6, 2008 Share Posted January 6, 2008 please help. ( a tough one i know) Let's find out: #include <date.au3> $sMsg = "" For $y = 1000 To 1900 Step 100 $sMsg &= $y & ": " & _DateDiff("d", $y & "/02/28", $y & "/03/01") & @CRLF Next For $y = 1995 To 2009 $sMsg &= $y & ": " & _DateDiff("d", $y & "/02/28", $y & "/03/01") & @CRLF Next MsgBox(64, "Leapyears?", $sMsg) The script shows the number of days from February 28 to Mach 01 in each year given. Note the pattern in even century years. A century year (1900 or 2000) is divisible by 4 and would be a leap year, but they are NOT... unless... they are divisible by 400. This is correct. Valuater's AutoIt 1-2-3, Class... Is now in Session!For those who want somebody to write the script for them: RentACoder"Any technology distinguishable from magic is insufficiently advanced." -- Geek's corollary to Clarke's law Link to comment Share on other sites More sharing options...
Aceguy Posted January 6, 2008 Author Share Posted January 6, 2008 WOW HOLY SHIT, ty. [u]My Projects.[/u]Launcher - not just for games & Apps (Mp3's & Network Files)Mp3 File RenamerMy File Backup UtilityFFXI - Realtime to Vana time Clock Link to comment Share on other sites More sharing options...
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