math: gregorian date to day number?

not acceptable: julian day number calculations.

rules are provided here:http://www.sizes.com/time/cal_gregorian.htm

you can code in javascript, c, or autoit, or maybe pseudocode.

this is not for a college project. I am stumped trying to produce appropriate math to convert gregorian day number to date and back.

I have this:

func GregorianToDtime($Y,$Mo,$D,$H,$Mi,$S,$ms, byref $dtime) ;do we have to offset these by 1?

;-----function good to 100,000 years if I coded it right.

;-----we will allow for a zero year.

;-----given gregorian date, convert to gregorian day number, then to dtime.

local $days= mod($Y,4)*365*$Y ;Every year whose number is not divisible by 4 without remainder, consists of 365 days;

;$days+=ceil((mod($Y,4) * mod($Y,100)>0)) * 366 * $Y ;this doesn't end up 1 when any mods are zeros! ;every year which is so divisible, but is not divisible by 100, of 366;

;$days+=((mod($Y,4) * mod($Y,100)>0)?1:0) * 366 * $Y ;this doesn't end up 1 when any mods are zeros! ;every year which is so divisible, but is not divisible by 100, of 366;

;this language doesn't have a ternary operator,so we implement as an if statement

local $temp=0

if (mod($Y,4) * mod($Y,100)>0) then ;this doesn't end up 1 when any mods are zeros!

$temp=1

endif

$days += $temp * 366 * $Y ;every year which is so divisible, but is not divisible by 100, of 366;

$days += (int($Y/100) - int($Y/400)) * 366; every year divisible by 100 but not by 400, again of 366.

$days += int($Y/4000) * 365 ;years divisible by 4000 should consist of 365 days.

$days += NumDaysInMonth( $Y,$Mo)

$days += $D

$days -= 1 ;subtract 1 day so days start at 0

$dtime = ($days * 1000*60*60*24) + ($H*1000*60*60) + ($Mi*60*1000) + ($S*1000) + $ms

return $dtime

endfunc

;http://www.sizes.com/time/cal_gregorian.htm

func DtimeToGregorian($dtime,byref $Y,byref $Mo,byref $D,byref $H,byref $Mi,byref $S,byref $ms) ;do we have to offset these by 1?

;-----function good to 100,000 years if I coded it right.

$ms=Mod($dtime,1000)

$S=Mod(Int($dtime / 1000),60)

$Mi=Mod(Int($dtime / (1000*60)),60)

$H=Mod(Int($dtime / (1000*60*60)),24)

local $days = int($dtime / (1000*60*60*24))

$days+=1 ;required for conversion due to offset of 1 day in gregorian calendar system

;-----determine the year, it's the biggest number

local $year=0, $year365=int($days/365), $year366=int($days/366)

if (mod($year366,100) * mod($year366,400) > 0 or mod($year365,4) * mod($year366,100) > 0) then

$Y = int($days/366)

$days -= ($Y*366)

elseif (mod($year365,4) > 0 or mod($year365, 4000)>0) then

$Y = int($days/365)

$days -= ($Y*365)

endif

local $i,$monthdays

for $Mo=12 to 1 step -1

$monthdays = SumOfDaysUpTo1stDayOfMonthWithoutYear( $Y,$Mo)

if ($monthdays < $days) then

;-----match!

$days-=NumDaysInMonth($Y,$Mo)

$D=$days

exitloop

endif

next

;-----return an array structure containing the resulting DTime

local $a[8]

$a[0]=7

$a[1]=$Y

$a[2]=$Mo

$a[3]=$D

$a[4]=$H

$a[5]=$Mi

$a[6]=$S

$a[7]=$ms

return $a

endfunc

func LeapIndex($Y)

if (0<>mod($Y, 4)) then

return 0;

elseif (0<>mod($Y, 100)) then

return 1;

elseif (0<>mod($Y, 400)) then

return 0;

else

return 1;

endif

endfunc

func NumDaysInMonth($Y,$Mo)

switch $mo

case 1,8,12

return 31

case 2

return LeapIndex($Y) + 28

case 3,5,7,10

return 31

case 4,6,9,11

return 30

endswitch

endfunc

func SumOfDaysUpTo1stDayOfMonthWithoutYear( $Y,$Mo)

;this is the sum of days up to 1st day of $Mo

local $sum=0,$i

for $i = 1 to $Mo-1

$sum += NumDaysInMonth($Y,$Mo)

next

return $sum+1

endfunc