convert algorithm to matlab code

3 visualizaciones (últimos 30 días)
Josaiah Ang
Josaiah Ang el 24 de Mayo de 2015
Comentada: Walter Roberson el 19 de Ag. de 2021
hi,
how do i write a matlab code to display armstrong number between two number input by user
Thank you.
Interger :: a, b, c
Interger :: abc, a3b3c3
Interger ::count
count =0
DO a = 0, 9
DO b = 0, 9
DO c = 0, 9
abc= a*100 + b*10 + c
a3b3c3 = a*3 + b*3 + c*3 !
IF (abc == a3b3c3) THEN
count = count + 1
WRITE(',') 'Armstrong number ', Count, ': ', abc
End If
End DO
End DO
End DO
  2 comentarios
Dilsher Mzuri
Dilsher Mzuri el 13 de Oct. de 2018
Editada: Walter Roberson el 15 de Oct. de 2018
hi everyone :
1- start
2- read image(a)
3- r=a
4- (n,k)=size(a);L=n*k
5- num=number of times for each pixel
6- pdf=num/(L-1)
7- cdf=sum(pdf)
8- s=sum((L-1)*cdf)
8- m=s+r
9- end
Algorithm of EHE method
the algorithm above is converted to matlab code
best rregards
Walter Roberson
Walter Roberson el 15 de Oct. de 2018
Dilsher Mzuri: I am not sure if that is a question or a response?

Iniciar sesión para comentar.

Respuestas (3)

Walter Roberson
Walter Roberson el 24 de Mayo de 2015

sunnia ikram
sunnia ikram el 28 de En. de 2021
please send me code of A* algorithem in matlab....for obstacle detection
  1 comentario
Walter Roberson
Walter Roberson el 28 de En. de 2021
That is not an Answer to the question.

Iniciar sesión para comentar.


Venkata Naga Sai Krishna Immadisetty
clc
clear all
clear canvas
n = -10:10;
figure(2);
a_values = [0.9,1.1,-0.9,-1.1];
for i=1:length(a_values)
fun = a_values(i).^n;
subplot(2,2,i)
stem(n,fun);
xlabel("Time samples")
ylabel("Amplitute")
title("a^n for a = (19BEC0668)"+a_values(i))
end
  1 comentario
Walter Roberson
Walter Roberson el 19 de Ag. de 2021
Please explain how this algorithm finds 3-digit Armstrong Numbers ?

Iniciar sesión para comentar.

Categorías

Más información sobre MATLAB en Help Center y File Exchange.

Etiquetas

Community Treasure Hunt

Find the treasures in MATLAB Central and discover how the community can help you!

Start Hunting!

Translated by