How to resolve the algorithm Square but not cube step by step in the REXX programming language
Published on 12 May 2024 09:40 PM
How to resolve the algorithm Square but not cube step by step in the REXX programming language
Table of Contents
Problem Statement
Show the first 30 positive integers which are squares but not cubes of such integers. Optionally, show also the first 3 positive integers which are both squares and cubes, and mark them as such.
Let's start with the solution:
Step by Step solution about How to resolve the algorithm Square but not cube step by step in the REXX programming language
Source code in the rexx programming language
/*REXX pgm shows N ints>0 that are squares and not cubes, & which are squares and cubes.*/
numeric digits 20 /*ensure handling of larger numbers. */
parse arg N . /*obtain optional argument from the CL.*/
if N=='' | N=="," then N= 30 /*Not specified? Then use the default.*/
sqcb= N<0 /*N negative? Then show squares & cubes*/
N = abs(N) /*define N to be the absolute value. */
w= (length(N) + 3) * 3 /*W: used for aligning output columns.*/
say ' count ' /*display the 1st line of the title. */
say ' ─────── ' /* " " 2nd " " " " */
@.= 0 /*@: stemmed array for computed cubes.*/
#= 0; ##= 0 /*count (integer): squares & not cubes.*/
do j=1 until #==N | ##==N /*loop 'til enough " " " " */
sq= j*j; cube= sq*j; @.cube= 1 /*compute the square of J and the cube.*/
if @.sq then do
##= ## + 1 /*bump the counter of squares & cubes. */
if \sqcb then counter= left('', 12) /*don't show this counter.*/
else counter= center(##, 12) /* do " " " */
say counter right(commas(sq), w) 'is a square and a cube'
end
else do
if sqcb then iterate
#= # + 1 /*bump the counter of squares & ¬ cubes*/
say center(#, 12) right(commas(sq), w) 'is a square and not a cube'
end
end /*j*/
exit 0 /*stick a fork in it, we're all done. */
/*──────────────────────────────────────────────────────────────────────────────────────*/
commas: parse arg ?; do jc=length(?)-3 to 1 by -3; ?=insert(',', ?, jc); end; return ?
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