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I have an idea, which is to take the remainder... divide by 365, the remainder Y1, then first take the remainder of Y1 divided by 4 (for leap years) to see how many leap years there are, and subtract several 1s from Y1 to get Y2. Then convert Y2 into months and days... (Haha, still execute is quite lazy)

Oh. Looked carefully. Indeed considered leap years. .

Principle research in progress?

Originally posted by slore at 2007-5-2 04:52 AM:
I have an idea, which is to use the remainder... divide by 365, the remainder Y1, then first take the remainder of Y1 divided by 4 (for leap years) to see how many leap years there are, and then subtract that number of 1s from Y1 to get Y2.

Then convert Y2 into months and days...
(Haha, still execute is quite lazy)

Now regardless of whether you are lazy or not...

Please come up with the code that everyone needs.

In this regard, please don't be lazy no matter what.

Oh, it seems like it's called New Script File.vbs. It seems to have been deleted. Then let's rewrite it, okay?

Originally posted by slore at 2007-5-2 05:28 AM:
Oh my god, it seems to be called New Script File.vbs

It seems to be deleted.. Then let's rewrite it = =

Hehe, thanks for your hard work:)

Looking forward to it.

Dim ADYear
Dim ADMonth
Dim ADDay
Dim ADDDay
Dim RnMod

Dim Mon(12)

Dim strDate, strADDay
strDate = "2007-3-4"
strADDay = 800

ADYear = Year(strDate)
RnMod = ADYear Mod 4
If RnMod = 0 Then tempRNYear = 1
ADMonth = Month(strDate)
ADDay = Day(strDate)

strADDay = strADDay + Date2Day(ADYear, ADMonth, ADDay)
tempYear = strADDay \ 365
tempRNday = strADDay Mod 365
If tempYear <> 0 Then
tempRNYear = tempRNYear + (RnMod + tempYear) \ 4
tempRNday = tempRNday - tempRNYear
End If

ADYear = ADYear + tempYear
MsgBox Day2Date(ADYear, tempRNday)


Function Date2Day(ByVal y, ByVal m, ByVal d)
Mon(1) = 31: Mon(2) = 28: Mon(3) = 31: Mon(4) = 30: Mon(5) = 31: Mon(6) = 30
Mon(7) = 31: Mon(8) = 31: Mon(9) = 30: Mon(10) = 31: Mon(11) = 30: Mon(12) = 31
Date2Day = 0

If IsDate(y & "/2/29") Then Mon(2) = 29
Dim i
For i = 1 To m - 1
Date2Day = Date2Day + Mon(i)
Next
Date2Day = Date2Day + d
End Function

Function Day2Date(ByVal y, ByVal Days)
Mon(2) = 28
If IsDate(y & "/2/29") Then Mon(2) = 29
Dim i

If Days > 31 Then
For i = 1 To 11
If Days < Mon(i) Then Exit For
Days = Days - Mon(i)
Next
ADMonth = i

Else
ADMonth = 1
End If
Day2Date = y & "-" & ADMonth & "-" & Days
End Function

Because there are 97 leap years every 400 years, there is a situation where there is a leap year every 8 years. So if 2000 is the initial year, don't calculate the days after 146000, because it will be off by 1 or 2 days, because I calculate based on a leap year every 4 years.

Here I only wrote the part of adding days... You can modify the part of subtracting days by yourself... I wrote this just to present my idea with code...

8000 days is also very fast... Hehe

Add below strADDay=800:
execute("msgbox #" & strDate & "#+" & strADDay) to verify...

10000 is still correct.. 2000000 is a mess... I won't change it... As long as the idea is clear, it's OK

Just tested it.. still relatively standard...

What I wrote is not enough.. I'll replace it with a new one later...

When organizing the index post, I found this post. Coincidentally, I posted a thread asking for the date N days ago yesterday. Please refer to here: Batch script to find the date N days ago

Originally posted by flyinspace at 2007-5-2 05:34 PM:

The algorithm is very simple.

Also smarter than mine... But because it didn't consider the existence of leap years.

There will be problems when calculating longer periods of days...

Adding a leap year judgment will be much better.

Didn't look closely at willsort's code;
But caught this line;

dd+y*365+y/4-y/100+y/400

This line shows that leap years are considered;
Isn't this usage a very classic algorithm?

A lot of code is not understandable.... Urgently studying...