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/* jd.c (conversions between calendar date and Julian day number) */
/***********************************************************************
* This code is part of GLPK (GNU Linear Programming Kit).
*
* Copyright (C) 2000-2013 Andrew Makhorin, Department for Applied
* Informatics, Moscow Aviation Institute, Moscow, Russia. All rights
* reserved. E-mail: <mao@gnu.org>.
*
* GLPK is free software: you can redistribute it and/or modify it
* under the terms of the GNU General Public License as published by
* the Free Software Foundation, either version 3 of the License, or
* (at your option) any later version.
*
* GLPK is distributed in the hope that it will be useful, but WITHOUT
* ANY WARRANTY; without even the implied warranty of MERCHANTABILITY
* or FITNESS FOR A PARTICULAR PURPOSE. See the GNU General Public
* License for more details.
*
* You should have received a copy of the GNU General Public License
* along with GLPK. If not, see <http://www.gnu.org/licenses/>.
***********************************************************************/
#include <stddef.h>
#include "jd.h"
/***********************************************************************
* NAME
*
* jday - convert calendar date to Julian day number
*
* SYNOPSIS
*
* #include "jd.h"
* int jday(int d, int m, int y);
*
* DESCRIPTION
*
* The routine jday converts a calendar date, Gregorian calendar, to
* corresponding Julian day number j.
*
* From the given day d, month m, and year y, the Julian day number j
* is computed without using tables.
*
* The routine is valid for 1 <= y <= 4000.
*
* RETURNS
*
* The routine jday returns the Julian day number, or negative value if
* the specified date is incorrect.
*
* REFERENCES
*
* R. G. Tantzen, Algorithm 199: conversions between calendar date and
* Julian day number, Communications of the ACM, vol. 6, no. 8, p. 444,
* Aug. 1963. */
int jday(int d, int m, int y)
{ int c, ya, j, dd;
if (!(1 <= d && d <= 31 &&
1 <= m && m <= 12 &&
1 <= y && y <= 4000))
return -1;
if (m >= 3)
m -= 3;
else
m += 9, y--;
c = y / 100;
ya = y - 100 * c;
j = (146097 * c) / 4 + (1461 * ya) / 4 + (153 * m + 2) / 5 + d +
1721119;
jdate(j, &dd, NULL, NULL);
if (d != dd)
return -1;
return j;
}
/***********************************************************************
* NAME
*
* jdate - convert Julian day number to calendar date
*
* SYNOPSIS
*
* #include "jd.h"
* int jdate(int j, int *d, int *m, int *y);
*
* DESCRIPTION
*
* The routine jdate converts a Julian day number j to corresponding
* calendar date, Gregorian calendar.
*
* The day d, month m, and year y are computed without using tables and
* stored in corresponding locations.
*
* The routine is valid for 1721426 <= j <= 3182395.
*
* RETURNS
*
* If the conversion is successful, the routine returns zero, otherwise
* non-zero.
*
* REFERENCES
*
* R. G. Tantzen, Algorithm 199: conversions between calendar date and
* Julian day number, Communications of the ACM, vol. 6, no. 8, p. 444,
* Aug. 1963. */
int jdate(int j, int *d_, int *m_, int *y_)
{ int d, m, y;
if (!(1721426 <= j && j <= 3182395))
return 1;
j -= 1721119;
y = (4 * j - 1) / 146097;
j = (4 * j - 1) % 146097;
d = j / 4;
j = (4 * d + 3) / 1461;
d = (4 * d + 3) % 1461;
d = (d + 4) / 4;
m = (5 * d - 3) / 153;
d = (5 * d - 3) % 153;
d = (d + 5) / 5;
y = 100 * y + j;
if (m <= 9)
m += 3;
else m -= 9,
y++;
if (d_ != NULL) *d_ = d;
if (m_ != NULL) *m_ = m;
if (y_ != NULL) *y_ = y;
return 0;
}
#ifdef GLP_TEST
#include <assert.h>
#include <stdio.h>
#include <stdlib.h>
int main(void)
{ int jbeg, jend, j, d, m, y;
jbeg = jday(1, 1, 1);
jend = jday(31, 12, 4000);
for (j = jbeg; j <= jend; j++)
{ assert(jdate(j, &d, &m, &y) == 0);
assert(jday(d, m, y) == j);
}
printf("Routines jday and jdate work correctly.\n");
return 0;
}
#endif
/* eof */
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