How to resolve the algorithm Proper divisors step by step in the PureBasic programming language

Published on 12 May 2024 09:40 PM
#Purebasic

How to resolve the algorithm Proper divisors step by step in the PureBasic programming language

Table of Contents

Problem Statement

The   proper divisors   of a positive integer N are those numbers, other than N itself, that divide N without remainder. For N > 1 they will always include 1,   but for N == 1 there are no proper divisors.

The proper divisors of     6     are   1, 2, and 3. The proper divisors of   100   are   1, 2, 4, 5, 10, 20, 25, and 50.

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Step by Step solution about How to resolve the algorithm Proper divisors step by step in the PureBasic programming language

Source code in the purebasic programming language

EnableExplicit

Procedure ListProperDivisors(Number, List Lst())
  If Number < 2 : ProcedureReturn : EndIf
  Protected i
  For i = 1 To Number / 2
    If Number % i = 0
      AddElement(Lst())
      Lst() = i
    EndIf
  Next
EndProcedure

Procedure.i CountProperDivisors(Number)
  If Number < 2 : ProcedureReturn 0 : EndIf
  Protected i, count = 0
  For i = 1 To Number / 2
    If Number % i = 0
      count + 1
    EndIf
  Next
  ProcedureReturn count
EndProcedure
  
Define n, count, most = 1, maxCount = 0
If OpenConsole()
  PrintN("The proper divisors of the following numbers are : ")
  PrintN("")
  NewList lst()
  For n = 1 To 10
    ListProperDivisors(n, lst())
    Print(RSet(Str(n), 3) + " -> ")
    If ListSize(lst()) = 0
      Print("(None)")
    Else  
      ForEach lst()
        Print(Str(lst()) + " ")
      Next
    EndIf
    ClearList(lst())
    PrintN("")
  Next
  For n = 2 To 20000
    count = CountProperDivisors(n)
    If count > maxCount
      maxCount = count
      most = n
    EndIf
  Next
  PrintN("")
  PrintN(Str(most) + " has the most proper divisors, namely " + Str(maxCount))
  PrintN("")
  PrintN("Press any key to close the console")
  Repeat: Delay(10) : Until Inkey() <> ""
  CloseConsole()
EndIf

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