Five weekends: Difference between revisions

The month of October in 2010 has five Fridays, five Saturdays, and five Sundays.

 

# Write a program to show all months that have this same characteristic of five full weekends from the year 1900 through 2100 (Gregorian calendar).

# Show the ''number'' of months with this property (there should be 201).

# Show at least the first and last five dates, in order.

 

'''Algorithm suggestions'''

 

'''Extra credit'''

 

Count and/or show all of the years which do not have at least one five-weekend month (there should be 29).

=={{header|Ada}}==

with Ada.Text_IO; use Ada.Text_IO;

with Ada.Calendar.Formatting; use Ada.Calendar;

Put_Line ("Number of months:" & Integer'Image (Months));

end Five_Weekends;

<pre style="height:30ex;overflow:scroll">

1901 3

</pre>

 

 

=={{header|AppleScript}}==

set noFiveWeekendYears to {}

 

set y to y + 12

end

 

<pre>Months with five weekends (201):

1901-March, 1902-August, 1903-May, 1904-January, 1904-July, ...

1984, 1990, 2001, 2007, 2012, 2018, 2029, 2035, 2040, 2046, 2057, 2063

2068, 2074, 2085, 2091, 2096</pre>

 

=={{header|AutoHotkey}}==

End_Year := 2100

31_Day_Months = 01,03,05,07,08,10,12

Years with no 5 day weekends between 1900 and 2100 : %No_5_Weekend_Count%

%All_Years_Without_5_Weekend%

 

<pre>

...

</pre>

 

=={{header|C}}==

#include <time.h>

 

printf("%d total\n", n);

return 0;

<pre>

Error: 1900 January

 

Not your typical method. Requires <code>ncal</code>.

#include <string.h>

 

 

return 0;

<pre>

1901-03 1902-08 1903-05 1904-01 1904-07 1905-12 1907-03 1908-05 1909-01 1909-10 1910-07 1911-12 1912-03 1913-08 1914-05 1915-01

Total: 201

</pre>

 

=={{header|C++}}==

{{libheader|Boost}}

#include <boost/date_time/gregorian/gregorian.hpp>

#include <algorithm>

std::cout << std::endl ;

return 0 ;

<pre style="height:30ex;overflow:scroll">

Between 1900 and 2100 , there are 201 months with 5 complete weekends!

</pre>

 

 

 

 

 

 

}

<pre>

</pre>

 

=={{header|Clojure}}==

java.text.DateFormatSymbols)

 

(println month "-" year))

count

 

=={{header|D}}==

 

// adjust to 1st day

for (Date when = Date(start.year, start.month, 1);

when < end;

return res;

}

 

}

 

void main() {

writeln("There are ", m.length,

" months of which the first and last five are:");

foreach (d; m[0 .. 5] ~ m[$ - 5 .. $])

 

<pre>There are 201 months of which the first and last five are:

 

=={{header|Delphi}}==

 

{$APPTYPE CONSOLE}

Writeln(Format('Months with 5 weekends: %d', [lFiveWeekendCount]));

Writeln(Format('Years with no 5 weekend months: %d', [lYearsWithout]));

 

=={{header|Erlang}}==

Two examples, both based on first building a nested list-of-lists like [Year1, Year2, ..., YearN], where each year is a sublist of year and month tuples, like [{YearN, 1}, {YearN, 2}, ..., {YearN, 12}].<br>

First, a pretty compact example intended for use with <b>escript</b>:

#!/usr/bin/env escript

%%

has_five({_Year, _Month}, _DaysNot31) ->

false.

Second, a more verbose Erlang module:

-module(five_weekends).

 

find_years_without_5w_months(List) ->

[Y || {[{Y,_M}|_], 0} <- List].

<pre>

There are 201 months with five full weekends.

2091

2096

</pre>

 

 

Using Zeller's congruence

implicit none

 

end function Day_of_week

<pre>Mar 1901

Aug 1902

Number of months with five weekends between 1900 and 2100 = 201

Number of years between 1900 and 2100 with no five weekend months = 29</pre>

 

=={{header|GAP}}==

# first is the list of months with five weekends between years y1 and y2 (included)

# second is the list of years without such months, in the same interval

# [ "Mar-1901", "Aug-1902", "May-1903", "Jan-1904", "Jul-1904" ]

r[1]{[n-4 .. n]};

=={{header|Go}}==

Using second algorithm suggestion:

 

import (

}

fmt.Println(len(haveNone), "total")

<pre>

Months with five weekends:

29 total

</pre>

 

=={{header|Inform 7}}==

Inform 7 has no built-in date functions, so this solution implements date types and a day-of-week function.

 

 

When play begins:

repeat with M running through months:

if M of Y is a happy month, decide no;

 

<pre>March 1901.

August 1902.

October 2100.

Found 201 months with five weekends and 29 years with no such months.</pre>

 

=={{header|J}}==

find5wkdMonths=: verb define

years=. range 2{. y

m5w=. (#~ 0 = weekday) >,{years;months;31 NB. 5 full weekends iff 31st is Sunday(0)

>'MMM YYYY' fmtDate toDayNo m5w

'''Usage:'''

201

(5&{. , '...' , _5&{.) find5wkdMonths 1900 2100 NB. First and last 5 months found

Oct 2100

# (range -. {:"1@(_ ". find5wkdMonths)) 1900 2100 NB. number of years without 5 weekend months

 

=={{header|Java}}==

In Java 1.5+ you can add <code>import static java.util.Calendar.*;</code> to the list of imports and remove all other occurrences of <code>Calendar.</code> from the rest of the code (e.g. <code>Calendar.FRIDAY</code> would simply become <code>FRIDAY</code>). It's more portable (and probably more clear) to leave the <code>Calendar.</code>'s in.

import java.util.GregorianCalendar;

 

public class FiveFSS {

private static boolean[] years = new boolean[201];

private static int[] month31 = {Calendar.JANUARY, Calendar.MARCH, Calendar.MAY,

Calendar.JULY, Calendar.AUGUST, Calendar.OCTOBER, Calendar.DECEMBER};

}

}

<pre style="height:30ex;overflow:scroll"> There are 201 months with five weekends from 1900 through 2100:

3-1901

 

=={{header|JavaScript}}==

{

// 0 is Sunday, 1 is Monday, ... 5 is Friday, 6 is Saturday

document.write('Years with no five-weekend months: ' + yearsWithoutFiveWeekends.length + '<br>');

}

 

<pre>1901 2

1902 7

1900,1906,1917,1923,1928,1934,1945,1951,1956,1962,1973,1979,1984,1990,2001,2007,2012,2018,2029,2035,2040,2046,2057,2063,2068,2074,2085,2091,2096

Years with no five-weekend months: 29</pre>

 

=={{header|k}}==

cal_j:(_jd[19000101]+!(-/_jd 21010101 19000101)) / enumerate the calendar

is_we:(cal_j!7) _lin 4 5 6 / identify friday saturdays and sundays

yn5:y@&~y _lin y5 / find any years not in the 5 weekend month list

`0:,"There are ",($#yn5)," years without any five-weekend months"

<pre>

There are 201 months with five weekends

1900,1906,1917,1923,1928,1934,1945,1951,1956,1962,1973,1979,1984,1990,2001,2007,2012,2018,2029,2035,2040,2046,2057,2063,2068,2074,2085,2091,2096</pre>

 

result = Select[Tuples[{Range@@years, months}], (DateString[# ~ Join ~ 1, "DayNameShort"] == "Fri")&];

 

Print["Last months: " , DateString[#,{"MonthName"," ","Year"}]& /@ result[[-5 ;; All]]];

Print[# // Length, " years without 5 weekend months:\n", #] &@

<pre>201 months with 5 weekends

First months: {March 1901, August 1902, May 1903, January 1904, July 1904}

1984, 1990, 2001, 2007, 2012, 2018, 2029, 2035, 2040, 2046, 2057, 2063,

2068, 2074, 2085, 2091, 2096}</pre>

 

=={{header|MUMPS}}==

{{libheader|VA Kernel|22.0}}

FIVE

;List and count the months between 1/1900 and 12/2100 that have 5 full weekends

. S:'N L=L_$S($L(L):",",1:"")_(R+1700)

W !,"For a total of "_C_" months.",!!,"There are "_$L(L,",")_" years with no five full weekends in any month.",!,"They are: "_L

USER>d ^FIVE

 

There are 29 years with no five full weekends in any month.

They are: 1900,1906,1917,1923,1928,1934,1945,1951,1956,1962,1973,1979,1984,1990,2001,2007,2012,2018,2029,2035,2040,2046,2057,2063,2068,2074,2085,2091,2096</pre>