speeding up my for loop
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%My challenge is to find the number of divisors for a number without using "divisors" inbuild function
%this program will work but it's taking too much time for big numbers
%I read about vectorization to reduce the time but found stucked
%help!!!
function y=divisors1(N)
sum=0;
for i=1:floor(N/2)
if lcm(N,i)==N
sum=sum+1;
end
end
y=1+sum;
Respuesta aceptada
Más respuestas (1)
Alan Stevens
el 27 de Feb. de 2021
Is this any quicker?
function y = divisors1(N)
i = 1:floor(N/2);
L = lcm(N,i);
y = sum(L==N) + 1;
end
1 comentario
Abhishek Sharma
el 27 de Feb. de 2021
Categorías
Más información sobre Loops and Conditional Statements en Centro de ayuda y File Exchange.
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