Richard A. Johnson

#include <iostream>
#include "boost/date_time/gregorian/gregorian.hpp"
 
int main(void)
{
    using namespace boost::gregorian;
    int count = 0;
    date_period dp(date(1901, Jan, 1), date(2000, Dec, 31));
    day_iterator iter(dp.begin());
 
    while (iter != dp.end())
    {
        if (iter->day() == 1 && iter->day_of_week().as_enum == 0)
            count++;
        ++iter;
    }
 
    std::cout << count << std::endl;
    return 0;
}

Responses

1. depesz says:
   

   December 27th, 2007 at 04:29:41 (#)

   time “miliseconds” doesn’t look impressive at all.

   on my desktop machine (dual code intel 1.8ghz, 2gb ram) i can find it in 5ms using *SQL*:

   # select count(*)
   =# from (
   =# select
   =# cast(start.date + ’1 month’::INTERVAL * generate_series(0, (finish.date – start.date)/28) as date) as date,
   =# finish.date as “finish”
   =# from
   =# (select ’2000-12-31′::date as date) as finish, (select ’1901-01-01′::date as date) as start
   =# ) x
   =# where 1 = extract(day from x.date) and 0 = extract(dow from x.date) AND x.date <= x.finish
   =# ;
   count
   ——-
   171
   (1 row)

   Time: 4.953 ms

   this is postgresql.

   i would hope that boosting library to be at least an order of magnitude faster.

2. </depesz> » Blog Archive » how many 1sts of any month were sundays - since 1901-01-01? says:
   

   December 27th, 2007 at 04:58:02 (#)

   [...] wrote about boost library for [...]

3. nixternal says:
   

   December 27th, 2007 at 11:49:08 (#)

   Problem 19 Answer: 171
   0.02 s

   This is a Celeron M 420 (1.6GHz) w/ 1.5GB of RAM. Better than the 1s I was getting with other library implementations for sure, and definitely a heck of a lot less code.

4. Fredrik says:
   

   December 27th, 2007 at 19:57:24 (#)

   Just had to try it in Python…

   
from datetime import date,timedelta
delta = timedelta(1)
d,end = date(1901, 1, 1),date(2000, 12, 31)
sundays = 0
while d < end:
if d.day == 1 and d.weekday() == 6:
sundays +=1
d += delta
print sundays


   I tried to measure the time and it varies between 15-50ms (AMD 64 4200+ win XP).

5. Yacin says:
   

   December 27th, 2007 at 20:18:49 (#)

   Psh, bash 4tw:

   for i in `seq 1 12` ; do for j in `seq 1901 2000`; do cal $i $j | grep " 1 2 3 4 5 6 7"; done; done | wc -l

6. Yacin says:
   

   December 27th, 2007 at 22:45:21 (#)

   for i in `seq 1 12` ; do for j in `seq 1901 2000`; do cal $i $j | grep " 1 2 3 4 5 6 7"; done; done | wc -l

   fixt

7. Yacin says:
   

   December 27th, 2007 at 23:04:23 (#)

   Ugh, it messes up the spaces, this one actually works

   for i in `seq 1 12` ; do for j in `seq 1901 2000`; do cal $i $j | sed ‘s/ //g’ | grep 1234567 ; done; done | wc -l

8. depesz says:
   

   December 28th, 2007 at 06:12:04 (#)

   this version of shell should be a bit faster:

   for a in `seq 1901 2000`; do cal $a; done | sed ‘s/ /\n/g’ | grep -c ‘^ 1 2 3 4 5 6 7′

9. depesz says:
   

   December 28th, 2007 at 06:54:54 (#)

   hmm – there should be *3* spaces in sed. i dont know how to add them in comment box. but it should be like:
   sed ‘s/XXX/\n/g’
   where “X” is space.

10. RichB says:
    

    December 29th, 2007 at 09:06:02 (#)

    JavaScript version using Zeller’s Congruence (doesn’t anyone read the comp.lang.c FAQ nowadays?). I haven’t timed it, but it’s on the order of milliseconds:

    window.Zeller={};

    Zeller.getDayOfWeek = function(y, m, d) {
    var t = [0,3,2,5,0,3,5,1,4,6,2,4];
    y-=m<3;
    return (y+Math.floor(y/4)-Math.floor(y/100)+Math.floor(y/400)+t[m-1]+d)%7;
    }

    var sundayCount=0;

    for(var i=1901; i<=2000; i++) {
    for(var j=1; j<=12; j++) {
    if(Zeller.getDayOfWeek(i, j, 1)==0) {
    sundayCount++;
    }
    }
    }
    document.write(“Number of sundays between 1901-01-01 and 2000-12-31 is ” + sundayCount);

11. RichB says:
    

    December 29th, 2007 at 09:24:21 (#)

    Correction: That’s not a Zellers Congruence algorithm. It’s from Tomohiko Sakamoto – see comp.lang.c faq for details.

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