How to resolve the algorithm Count in factors step by step in the ALGOL 68 programming language

Published on 12 May 2024 09:40 PM
#Algol 68

How to resolve the algorithm Count in factors step by step in the ALGOL 68 programming language

Table of Contents

Problem Statement

Write a program which counts up from   1,   displaying each number as the multiplication of its prime factors. For the purpose of this task,   1   (unity)   may be shown as itself.

2   is prime,   so it would be shown as itself.       6   is not prime;   it would be shown as

2 × 3

{\displaystyle 2\times 3}

. 2144   is not prime;   it would be shown as

2 × 2 × 2 × 2 × 2 × 67

{\displaystyle 2\times 2\times 2\times 2\times 2\times 67}

.

Let's start with the solution:

Step by Step solution about How to resolve the algorithm Count in factors step by step in the ALGOL 68 programming language

Source code in the algol programming language

OP +:= = (REF FLEX []INT a, INT b) VOID:
   BEGIN
      [UPB a + 1] INT c;
      c[:UPB a] := a;
      c[UPB a+1:] := b;
      a := c
   END;


PROC factorize = (INT nn) []INT:
   BEGIN
      IF nn = 1 THEN (1)
      ELSE
	  INT k := 2, n := nn;
	  FLEX[0]INT result;
	  WHILE n > 1 DO
	      WHILE n MOD k = 0 DO
		  result +:= k;
		  n := n % k
	      OD;
	      k +:= 1
	  OD;
	  result
      FI 
   END;
 
FLEX[0]INT factors;
FOR i TO 22 DO
    factors := factorize (i);
    print ((whole (i, 0), " = "));
    FOR j TO UPB factors DO
       (j /= 1 | print (" × "));
	print ((whole (factors[j], 0)))
    OD;
    print ((new line))
OD

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