How to resolve the algorithm 99 bottles of beer step by step in the Ra programming language
Published on 12 May 2024 09:40 PM
How to resolve the algorithm 99 bottles of beer step by step in the Ra programming language
Table of Contents
Problem Statement
Display the complete lyrics for the song: 99 Bottles of Beer on the Wall.
The lyrics follow this form: ... and so on, until reaching 0 (zero). Grammatical support for 1 bottle of beer is optional. As with any puzzle, try to do it in as creative/concise/comical a way as possible (simple, obvious solutions allowed, too).
Let's start with the solution:
Step by Step solution about How to resolve the algorithm 99 bottles of beer step by step in the Ra programming language
Source code in the ra programming language
class BottlesOfBeer
**Prints the "99 Bottles of Beer" song"**
on start
args := program arguments
# If no arguments given, print the song once
if args empty
.printSong
# Otherwise, print the song the given number of times
else
.printSong(integer.parse(args[0]))
shared
define printSong(times := 1)
**Print the song the given number of times**
for times, _printSong
define _printSong
**Print the song**
# Print for bottles 99 to 1
for bottle in 99 to 0 by -1
print "[_bottles(bottle)] of beer on the wall
[_bottles(bottle)] of beer
Take one down, pass it around
[_bottles(bottle - 1)] of beer on the wall
"
print "No more bottles of beer on the wall
No more bottles of beer
Go to the store, buy some more
99 bottles of beer on the wall"
define _bottles(bottle as integer) as String
**
If bottle is 0, returns "No more bottles"
If bottle is 1, returns "1 bottle"
Otherwise, returns "[bottle] bottles"
**
if bottle = 0, return "No more bottles"
if bottle = 1, return "1 bottle"
return "[bottle] bottles"
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