Calculate the current date.

Applications Programming
1

Give your an integer (e.g. 0x0076f676) representing the number of minutes elapsed since January 1, 1996.
How to calculate the current date which format should be "year-month-day-hour-minutes" ?

2

I'm not quite following this.

Are you saying you want to work out what the current year month and day is from the number of minutes elapsed since January 1, 1996?

3

This looked interesting, if a bit like homework. Tinker with this:

//Give your an integer (e.g. 0x0076f676) representing the number of minutes elapsed since January 1, 1996. 
//How to calculate the current date which format should be "year-month-day-hour-minutes" ?
// jan1.c
// usage:  ./jan1 0x0012345

#include <stdlib.h>
#include <stdio.h>
#include <time.h>
#include <strings.h>

time_t jan_1_1996(void)  // create artificial epoch seconds 
{
    struct tm j;
    j.tm_year= 1996 - 1900;
    j.tm_mon= 0;
    j.tm_mday=1;
    j.tm_hour=0;
    j.tm_min= 0;
    j.tm_sec= 0;
    j.tm_isdst=0;
    return (time_t)mktime(&j);
}

time_t
hex2dec(const char *hex)
{
    const char *p=hex;
    if(memcmp(p, "0x", 2)== 0) 
       p+=2;
    return (time_t) strtol(hex, (char **)0, 16);
}

void
get_date(const char *hex)
{
    time_t elapsed= jan_1_1996();
    char tmp[80]={0x0};
    elapsed+=( 60 * hex2dec(hex));
    strftime(tmp,80, "%Y-%m-%d-%H-%M", localtime(&elapsed) );
    printf("Corrected date is for %s: %s\n", hex, tmp);
}

int main(int argc, char **argv)
{
    if(argc==2)
      get_date(argv[1]);
    return 0;  
}

There is NO error checking in this code at all. Not meant for production systems.

4

If you don't care about what language to use, Perl would be much easier.

#!/usr/bin/perl

use POSIX;

$t = mktime(0, 0, 0, 1, 0, 96);
$i = 0x0076f676 * 60;
$s = strftime "%Y-%m-%d-%H-%M", localtime($t + $i);
print "$s\n";