speeding up my for loop
1 Ansicht (letzte 30 Tage)
Ältere Kommentare anzeigen
Abhishek Sharma
am 27 Feb. 2021
Kommentiert: Abhishek Sharma
am 27 Feb. 2021
%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;
0 Kommentare
Akzeptierte Antwort
Bruno Luong
am 27 Feb. 2021
Bearbeitet: Bruno Luong
am 27 Feb. 2021
function y=divisors2(N)
f = factor(N);
[~,~,J] = unique(f);
n = accumarray(J,1);
y = prod(n+1);
end
Test
>> N=27022021
N =
27022021
>> divisors2(N)
ans =
8
Weitere Antworten (1)
Alan Stevens
am 27 Feb. 2021
Is this any quicker?
function y = divisors1(N)
i = 1:floor(N/2);
L = lcm(N,i);
y = sum(L==N) + 1;
end
Siehe auch
Kategorien
Mehr zu Loops and Conditional Statements finden Sie in Help Center und File Exchange
Community Treasure Hunt
Find the treasures in MATLAB Central and discover how the community can help you!
Start Hunting!