1/*-
2 * Copyright (C) 1993-1996 by Andrey A. Chernov, Moscow, Russia.
3 * All rights reserved.
4 *
5 * Redistribution and use in source and binary forms, with or without
6 * modification, are permitted provided that the following conditions
7 * are met:
8 * 1. Redistributions of source code must retain the above copyright
9 *    notice, this list of conditions and the following disclaimer.
10 * 2. Redistributions in binary form must reproduce the above copyright
11 *    notice, this list of conditions and the following disclaimer in the
12 *    documentation and/or other materials provided with the distribution.
13 *
14 * THIS SOFTWARE IS PROVIDED BY THE AUTHOR ``AS IS'' AND
15 * ANY EXPRESS OR IMPLIED WARRANTIES, INCLUDING, BUT NOT LIMITED TO, THE
16 * IMPLIED WARRANTIES OF MERCHANTABILITY AND FITNESS FOR A PARTICULAR PURPOSE
17 * ARE DISCLAIMED.  IN NO EVENT SHALL THE REGENTS OR CONTRIBUTORS BE LIABLE
18 * FOR ANY DIRECT, INDIRECT, INCIDENTAL, SPECIAL, EXEMPLARY, OR CONSEQUENTIAL
19 * DAMAGES (INCLUDING, BUT NOT LIMITED TO, PROCUREMENT OF SUBSTITUTE GOODS
20 * OR SERVICES; LOSS OF USE, DATA, OR PROFITS; OR BUSINESS INTERRUPTION)
21 * HOWEVER CAUSED AND ON ANY THEORY OF LIABILITY, WHETHER IN CONTRACT, STRICT
22 * LIABILITY, OR TORT (INCLUDING NEGLIGENCE OR OTHERWISE) ARISING IN ANY WAY
23 * OUT OF THE USE OF THIS SOFTWARE, EVEN IF ADVISED OF THE POSSIBILITY OF
24 * SUCH DAMAGE.
25 */
26
27#include <sys/cdefs.h>
28__FBSDID("$FreeBSD$");
29
30#include <stdio.h>
31#include <stdlib.h>
32#include <string.h>
33
34#include "calendar.h"
35
36#define	PASKHA		"paskha"
37#define	PASKHALEN	(sizeof(PASKHA) - 1)
38
39/* return difference in days between Julian and Gregorian calendars */
40int
41j2g(int year)
42{
43	return (year < 1500) ?
44		0 :
45		10 + (year/100 - 16) - ((year/100 - 16) / 4);
46}
47
48/* return year day for Orthodox Easter using Gauss formula */
49/* (new style result) */
50
51int
52paskha(int R) /*year*/
53{
54	int a, b, c, d, e;
55	static int x = 15;
56	static int y = 6;
57	int *cumday;
58
59	a = R % 19;
60	b = R % 4;
61	c = R % 7;
62	d = (19 * a + x) % 30;
63	e = (2 * b + 4 * c + 6 * d + y) % 7;
64	cumday = cumdaytab[isleap(R)];
65	return (((cumday[3] + 1) + 22) + (d + e) + j2g(R));
66}
67