1 /*- 2 * SPDX-License-Identifier: BSD-2-Clause 3 * 4 * Copyright (C) 1993-1996 by Andrey A. Chernov, Moscow, Russia. 5 * All rights reserved. 6 * 7 * Redistribution and use in source and binary forms, with or without 8 * modification, are permitted provided that the following conditions 9 * are met: 10 * 1. Redistributions of source code must retain the above copyright 11 * notice, this list of conditions and the following disclaimer. 12 * 2. Redistributions in binary form must reproduce the above copyright 13 * notice, this list of conditions and the following disclaimer in the 14 * documentation and/or other materials provided with the distribution. 15 * 16 * THIS SOFTWARE IS PROVIDED BY THE AUTHOR ``AS IS'' AND 17 * ANY EXPRESS OR IMPLIED WARRANTIES, INCLUDING, BUT NOT LIMITED TO, THE 18 * IMPLIED WARRANTIES OF MERCHANTABILITY AND FITNESS FOR A PARTICULAR PURPOSE 19 * ARE DISCLAIMED. IN NO EVENT SHALL THE REGENTS OR CONTRIBUTORS BE LIABLE 20 * FOR ANY DIRECT, INDIRECT, INCIDENTAL, SPECIAL, EXEMPLARY, OR CONSEQUENTIAL 21 * DAMAGES (INCLUDING, BUT NOT LIMITED TO, PROCUREMENT OF SUBSTITUTE GOODS 22 * OR SERVICES; LOSS OF USE, DATA, OR PROFITS; OR BUSINESS INTERRUPTION) 23 * HOWEVER CAUSED AND ON ANY THEORY OF LIABILITY, WHETHER IN CONTRACT, STRICT 24 * LIABILITY, OR TORT (INCLUDING NEGLIGENCE OR OTHERWISE) ARISING IN ANY WAY 25 * OUT OF THE USE OF THIS SOFTWARE, EVEN IF ADVISED OF THE POSSIBILITY OF 26 * SUCH DAMAGE. 27 */ 28 29 #include <sys/cdefs.h> 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 */ 40 int 41 j2g(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 51 int 52 paskha(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