星期计算的小问题（一个世纪的基本世界末日）
从这个在线计算器：http://homer.freeshell.org/dd.cgi 使用其数据我已经成功编写了一个工作版本，但是它的数据仅限于 1500 到 2600 年。我想修改（并制作一个更好的版本），以便我可以计算任何年份 > 2600.
参考表X，实际上是否有一个公式可以计算所有基准世纪的基准末日 （2600以上）？
我自己尝试过将几个世纪设置得比这个高，例如2700给了我一个基本的世界末日“00”，2800给了“02;”，2900又回到了“00”......
感谢帮助。
From this online calculator: http://homer.freeshell.org/dd.cgi using its data I've successfully written a working version, however its data is limited to years 1500 to 2600. I want to modify (and make a better one) so that I can calculate for any year > 2600.
Referring to Table X, is there actually a formula to calculate the base doomsday for all base centuries (above 2600)?
I've tried working it out myself by putting centuries higher than this e.g. 2700 gave me a base doomsday of '00', 2800 gave '02;, 2900 back to '00' again...
Help appreciated.
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据我了解，该页面的“Base Doomsday”只是一个偏移量，以允许四-闰日计算的百年周期。因此，您只需添加四个世纪的块即可将其无限期地扩展到未来。
As I understand it, that page's “Base Doomsday” is just an offset to allow for the four-hundred-year cycle of leap day calculations. So, you can extend it indefinitely into the future simply by adding blocks of four centuries.
还有其他计算器可以做到这一点吗？
计算星期几的两种常用方法
给定的日期是世界末日，您正在使用该日期，
和 泽勒同余
www.merlyn.demon.co.uk 提供
关于日期/时间计算、各种日历的一些非常有趣的信息
系统和重要日期，因为它们与日历/日期计算相关。
Are there any other calculators out there that do this?
Two common methods for calculating the day of the week
given a date are Doomsday, which you are using,
and Zeller's Congruence
www.merlyn.demon.co.uk provides
some really interesting information on date/time calculations, various calendar
systems and significant dates as they relate to calendar/date calculations.
此链接中的计算器 http://homer.freeshell.org/dd.cgi 是最好的就为人类干净清晰地解释世界末日算法而言，有一点警告。
如果你输入2/29/1900，它会说一个星期四。嗯，没有 2/29/1900，因为它不是闰年。
当然，如果您输入 1/35/2016，它也会为您“垃圾进垃圾出”。
想象一下，一年只有 364 天，那么每个日期的星期几将永远不会年复一年地改变，因为
mod(364,7)==0。但我们一年有 365 天，所以每年这一天向前推进 1，这就是第二项
mod(year, 7)的由来。此外，每 4 年有一个闰年，这有助于最后一项
mod(year, 4)。但每 100 年，您减去一个闰年，每 400 年，您添加一个闰年。这就是第一个术语“3,2,0,5”的用武之地。
你看，这都是因为这个闰年和
mod(365,7)==1业务。7/11，5to9对记忆Z表有很大帮助。
The calculator at this link http://homer.freeshell.org/dd.cgi is the best in terms of explaining doomsday algorithm cleanly and clearly for human, with one little caveat.
If you input 2/29/1900, it would say it's a Thursday. Well, there is no 2/29/1900, because it's not a leap year.
Of course if your input 1/35/2016, it would "garbage-in-garbage-out" for you as well.
Imagine there are only 364 days in a year, then the day of week for each date will never change year after year, because
mod(364,7)==0.But we have 365 days a year, so the day steps forward 1 each year, that's where the second term
mod(year, 7)comes from.In addition, every 4 year, there is a leap year, which contributes to the last term
mod(year, 4).But every 100 years, you subtract a leap year, and every 400 years, you add one leap year. That's where the first term "3,2,0,5" comes in.
You see, it's all because of this leap year, and
mod(365,7)==1business.7/11, 5to9 helps to remember table Z greatly.
尝试一下公式：
因为它是显式公式。
例如：
Try the formula:
Because it's the explicit formula.
For example: