How to resolve the algorithm Five weekends step by step in the AutoHotkey programming language

Published on 12 May 2024 09:40 PM
#Autohotkey

How to resolve the algorithm Five weekends step by step in the AutoHotkey programming language

Table of Contents

Problem Statement

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

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).

Let's start with the solution:

Step by Step solution about How to resolve the algorithm Five weekends step by step in the AutoHotkey programming language

Source code in the autohotkey programming language

Year := 1900
End_Year := 2100
31_Day_Months = 01,03,05,07,08,10,12

While Year <= End_Year
  {
    Loop, Parse, 31_Day_Months, CSV
      {
        FormatTime, Day, %Year%%A_LoopField%01, dddd
        IfEqual, Day, Friday
          {
            All_Months_With_5_Weekends .=  A_LoopField . "/" . Year ",    "
            5_Weekend_Count++
            Year_Has_5_Weekend_Month := 1
          }
      }
   IfEqual, Year_Has_5_Weekend_Month, 0
      {
        All_Years_Without_5_Weekend .= Year ",   "
        No_5_Weekend_Count ++
      }
   Year ++
   Year_Has_5_Weekend_Month := 0
  }
; Trim the spaces and comma off the last item.
StringTrimRight, All_Months_With_5_Weekends, All_Months_With_5_Weekends, 5
StringTrimRight, All_Years_Without_5_Weekend, All_Years_Without_5_Weekend, 4
MsgBox,
(
Months with 5 day weekends between 1900 and 2100 : %5_Weekend_Count%
%All_Months_With_5_Weekends%
)
MsgBox,
(
Years with no 5 day weekends between 1900 and 2100 : %No_5_Weekend_Count%
%All_Years_Without_5_Weekend%
)

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