C中的日期推算法…………：》

[Meet03]

各位哥哥姐姐们好：
我有个问题帮解决一下，在C中怎样解决某年某月某日是星期几呢：》各位帮一帮忙：》

[zhanxiaozhang]

这里有个算法：
Contributor: EARL DUNOVANT

{
EARL DUNOVANT

> Which date is what day For a particular month.

Zeller's Congruence is an algorithm that calculates a day of the week given
a year, month and day. Created in 1887(!). Jeff Duntemann of PC Techniques
fame implemented it in TP in the 11/90 issue of Dr Dobbs Journal, With a
(115 min left), (H)elp, More? major kludge because TP's MOD operator returns a remainder instead of a
True mathematical modulus. I added the Kludge Alert banner that I use in my
own code.
}

Function CalcDayOfWeek(Year, Month, Day : Integer) : Integer;
Var
Century,
Holder : Integer;
begin
{ First test For error conditions on input values: }
if (Year < 0) or (Month < 1) or (Month > 12) or (Day < 1) or (Day > 31) then
CalcDayOfWeek := -1 { Return -1 to indicate an error }
else
{ Do the Zeller's Congruence calculation as Zeller himself }
{ described it in "Acta Mathematica" #7, Stockhold, 1887. }
begin
{ First we separate out the year and the century figures: }
Century := Year div 100;
Year := Year MOD 100;
{ Next we adjust the month such that March remains month #3, }
{ but that January and February are months #13 and #14, }
{ *but of the previous year*: }
if Month < 3 then
begin
Inc(Month, 12);
if Year > 0 then
Dec(Year, 1) { The year before 2000 is }
else { 1999, not 20-1... }
begin
Year := 99;
Dec(Century);
end;
end;

{ Here's Zeller's seminal black magic: }
Holder := Day; { Start With the day of month }
Holder := Holder + (((Month + 1) * 26) div 10); { Calc the increment }
Holder := Holder + Year; { Add in the year }
Holder := Holder + (Year div 4); { Correct For leap years }
Holder := Holder + (Century div 4); { Correct For century years }
Holder := Holder - Century - Century; { DON'T KNOW WHY HE DID THIS! }
{***********************KLUDGE ALERT!***************************}
While Holder < 0 do { Get negative values up into }
Inc(Holder, 7); { positive territory before }
{ taking the MOD... }
Holder := Holder MOD 7; { Divide by 7 but keep the }
{ remainder rather than the }
{ quotient }
{***********************KLUDGE ALERT!***************************}
{ Here we "wrap" Saturday around to be the last day: }
if Holder = 0 then
Holder := 7;

{ Zeller kept the Sunday = 1 origin; computer weenies prefer to }
{ start everything With 0, so here's a 20th century kludge: }
Dec(Holder);

CalcDayOfWeek := Holder; { Return the end product! }
end;
end;

[ahwang]

程序是什么语言的啊？
怎么没有主函数啊？
在C语言中有ENG吗？

[zhanxiaozhang]

这个是pascal

[Fever]

请到VC6的安装目录中找VC98\CRT\SRC\time.c和time.h
结构体tm中保存了日期和时间相关的信息，可用函数asctime来转成字符输出

[seaquester]

下面的代码利用蔡勒（Zeller）公式来计算星期（只适合于1582年10月15日之后的情形），在VC++ 。net 2003上测试通过：


#include <iostream>
using namespace std;

typedef unsigned long DWORD;

//-------------------------------------------------------------------------------
// 蔡勒（Zeller）公式（只适合于1582年10月15日之后的情形）：
// w=y+[y/4]+[c/4]-2c+[26(m+1)/10]+d-1
// 公式中的符号含义如下，
// w：星期；w对7取模得：0-星期日，1-星期一，2-星期二，3-星期三，4-星期四，
// 5-星期五，6-星期六
// c：世纪（年的高两位数）；
// y：年（年的低两位数）；
// m：月（m大于等于3，小于等于14，即在蔡勒公式中，某年的1、2月要看作上一年的13、14月
// 来计算，比如2005年1月1日要看作2004年的13月1日来计算）；
// d：日；
// []代表取整，即只要整数部分。
//
// 以2005年2月14日为例：c=20，y=4，m=14，d=14
// w = 4 + [4/4] + [20/4] - 2*20 + [26*(14+1)/10] + 14 - 1
// = 4 + 1 + 5 - 40 + 39 + 14 - 1
// = 22 (除以7余1)
// 所以2005年2月14日是星期一。
//
//-------------------------------------------------------------------------------
DWORD CalcDayofWeek(DWORD dwYear, DWORD dwMonth, DWORD dwDay)
{
DWORD dwWeek;
DWORD dwCentury;
DWORD dwNewYear, dwNewMonth;

// 某年的1、2月要看作上一年的13、14月
if (dwMonth < 3)
{
dwNewMonth = dwMonth + 12;
dwNewYear = (dwYear - 1)%100;
}
else
{
dwNewMonth = dwMonth;
dwNewYear = dwYear%100;
}

dwCentury = dwYear/100;

dwWeek = dwNewYear + dwNewYear/4 + dwCentury/4 - 2*dwCentury
+ 26*(dwNewMonth+1)/10 + dwDay - 1;
dwWeek = dwWeek % 7;

return dwWeek;
}


int main(int argc, char* argv[])
{
char szDay[25] = {32};

DWORD w = CalcDayofWeek(2005, 2, 14);

switch(w)
{
case 0:
sprintf(szDay, "Sunday");
break;
case 1:
sprintf(szDay, "Monday");
break;
case 2:
sprintf(szDay, "Tuesday");
break;
case 3:
sprintf(szDay, "Wednesday");
break;
case 4:
sprintf(szDay, "Thursday");
break;
case 5:
sprintf(szDay, "Friday");
break;
case 6:
sprintf(szDay, "Saturday");
break;
default:
sprintf(szDay, "Unknown");
break;
}
cout << szDay << endl;
return 0;
}

[Meet03]

我想用c写啊:>

[ybt631]

c的函数库<time.h>里面有这方面的函数。
查查一般编译器手册就可以了

[cyj2008]

以某年某月某日为参考日，假设参考日为星期k（k=0,1,2,...,6），其中k=0代表星期天。
那么要求任意一天为星期几，只需求出其与参考日所相差的天数elapsedDays,(elapsedDays+k) Mod 7即为结果

[Meet03]

谢谢：》