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Aceguy

_datediff

3 posts in this topic




:)

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.

:P


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

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