How to resolve the algorithm Abundant, deficient and perfect number classifications step by step in the Ada programming language
Published on 12 May 2024 09:40 PM
How to resolve the algorithm Abundant, deficient and perfect number classifications step by step in the Ada 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 Ada programming language
Source code in the ada programming language
with Ada.Text_IO, Generic_Divisors;
procedure ADB_Classification is
function Same(P: Positive) return Positive is (P);
package Divisor_Sum is new Generic_Divisors
(Result_Type => Natural, None => 0, One => Same, Add => "+");
type Class_Type is (Deficient, Perfect, Abundant);
function Class(D_Sum, N: Natural) return Class_Type is
(if D_Sum < N then Deficient
elsif D_Sum = N then Perfect
else Abundant);
Cls: Class_Type;
Results: array (Class_Type) of Natural := (others => 0);
package NIO is new Ada.Text_IO.Integer_IO(Natural);
package CIO is new Ada.Text_IO.Enumeration_IO(Class_Type);
begin
for N in 1 .. 20_000 loop
Cls := Class(Divisor_Sum.Process(N), N);
Results(Cls) := Results(Cls)+1;
end loop;
for Class in Results'Range loop
CIO.Put(Class, 12);
NIO.Put(Results(Class), 8);
Ada.Text_IO.New_Line;
end loop;
Ada.Text_IO.Put_Line("--------------------");
Ada.Text_IO.Put("Sum ");
NIO.Put(Results(Deficient)+Results(Perfect)+Results(Abundant), 8);
Ada.Text_IO.New_Line;
Ada.Text_IO.Put_Line("====================");
end ADB_Classification;
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