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1d386b48a5
Remove /^[\s*]*__FBSDID\("\$FreeBSD\$"\);?\s*\n/
67 lines
2.1 KiB
C
67 lines
2.1 KiB
C
/*-
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* SPDX-License-Identifier: BSD-2-Clause
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*
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* Copyright (C) 1993-1996 by Andrey A. Chernov, Moscow, Russia.
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* All rights reserved.
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*
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* Redistribution and use in source and binary forms, with or without
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* modification, are permitted provided that the following conditions
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* are met:
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* 1. Redistributions of source code must retain the above copyright
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* notice, this list of conditions and the following disclaimer.
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* 2. Redistributions in binary form must reproduce the above copyright
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* notice, this list of conditions and the following disclaimer in the
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* documentation and/or other materials provided with the distribution.
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*
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* THIS SOFTWARE IS PROVIDED BY THE AUTHOR ``AS IS'' AND
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* ANY EXPRESS OR IMPLIED WARRANTIES, INCLUDING, BUT NOT LIMITED TO, THE
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* IMPLIED WARRANTIES OF MERCHANTABILITY AND FITNESS FOR A PARTICULAR PURPOSE
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* ARE DISCLAIMED. IN NO EVENT SHALL THE REGENTS OR CONTRIBUTORS BE LIABLE
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* FOR ANY DIRECT, INDIRECT, INCIDENTAL, SPECIAL, EXEMPLARY, OR CONSEQUENTIAL
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* DAMAGES (INCLUDING, BUT NOT LIMITED TO, PROCUREMENT OF SUBSTITUTE GOODS
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* OR SERVICES; LOSS OF USE, DATA, OR PROFITS; OR BUSINESS INTERRUPTION)
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* HOWEVER CAUSED AND ON ANY THEORY OF LIABILITY, WHETHER IN CONTRACT, STRICT
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* LIABILITY, OR TORT (INCLUDING NEGLIGENCE OR OTHERWISE) ARISING IN ANY WAY
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* OUT OF THE USE OF THIS SOFTWARE, EVEN IF ADVISED OF THE POSSIBILITY OF
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* SUCH DAMAGE.
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*/
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#include <sys/cdefs.h>
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#include <stdio.h>
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#include <stdlib.h>
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#include <string.h>
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#include "calendar.h"
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#define PASKHA "paskha"
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#define PASKHALEN (sizeof(PASKHA) - 1)
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/* return difference in days between Julian and Gregorian calendars */
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int
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j2g(int year)
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{
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return (year < 1500) ?
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0 :
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10 + (year/100 - 16) - ((year/100 - 16) / 4);
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}
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/* return year day for Orthodox Easter using Gauss formula */
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/* (new style result) */
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int
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paskha(int R) /*year*/
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{
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int a, b, c, d, e;
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static int x = 15;
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static int y = 6;
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int *cumday;
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a = R % 19;
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b = R % 4;
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c = R % 7;
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d = (19 * a + x) % 30;
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e = (2 * b + 4 * c + 6 * d + y) % 7;
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cumday = cumdaytab[isleap(R)];
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return (((cumday[3] + 1) + 22) + (d + e) + j2g(R));
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}
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