How to resolve the algorithm Abundant, deficient and perfect number classifications step by step in the Action! programming language

Published on 12 May 2024 09:40 PM
#Action!

How to resolve the algorithm Abundant, deficient and perfect number classifications step by step in the Action! programming language

Table of Contents

Problem Statement

These define three classifications of positive integers based on their   proper divisors. Let   P(n)   be the sum of the proper divisors of   n   where the proper divisors are all positive divisors of   n   other than   n   itself.

6   has proper divisors of   1,   2,   and   3. 1 + 2 + 3 = 6,   so   6   is classed as a perfect number.

Calculate how many of the integers   1   to   20,000   (inclusive) are in each of the three classes. Show the results here.

Let's start with the solution:

Step by Step solution about How to resolve the algorithm Abundant, deficient and perfect number classifications step by step in the Action! programming language

Source code in the action! programming language

PROC FillSumOfDivisors(CARD ARRAY pds CARD size,maxNum,offset)
  CARD i,j

  FOR i=0 TO size-1
  DO
    pds(i)=1
  OD
  FOR i=2 TO maxNum DO
    FOR j=i+i TO maxNum STEP i
    DO
      IF j>=offset THEN
        pds(j-offset)==+i
      FI
    OD
  OD
RETURN

PROC Main()
  DEFINE MAXNUM="20000"
  DEFINE HALFNUM="10000"
  CARD ARRAY pds(HALFNUM+1)
  CARD def,perf,abud,i,sum,offset
  BYTE CRSINH=$02F0 ;Controls visibility of cursor
 
  CRSINH=1 ;hide cursor
  Put(125) PutE() ;clear the screen
  PrintE("Please wait...")

  def=1 perf=0 abud=0
  FillSumOfDivisors(pds,HALFNUM+1,HALFNUM,0)
  FOR i=2 TO HALFNUM
  DO
    sum=pds(i)
    IF sum
    ELSEIF sum=i THEN perf==+1
    ELSE abud==+1 FI
  OD

  offset=HALFNUM
  FillSumOfDivisors(pds,HALFNUM+1,MAXNUM,offset)
  FOR i=HALFNUM+1 TO MAXNUM
  DO
    sum=pds(i-offset)
    IF sum
    ELSEIF sum=i THEN perf==+1
    ELSE abud==+1 FI
  OD

  PrintF("  Numbers: %I%E",MAXNUM)
  PrintF("Deficient: %I%E",def)
  PrintF("  Perfect: %I%E",perf)
  PrintF("  Abudant: %I%E",abud)
RETURN

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