Re: Finding LCM & GCD in C
Lets see the performance.
:confused: Which is the fastest algorithm? :) GCM seems to be at most n :( LCM seems to be at most n*m :p so, if you do lcm(a,b) := (a*b)/gcm(a,b), you get a n complexity :eek: This is mathematics. 
Re: Finding LCM & GCD in C
Swamp is the subject of another post:
http://www.go4expert.com/showthread.php?t=1732 
Re: Finding LCM & GCD in C
Quote:

Re: Finding LCM & GCD in C
clocking and shabbir are speaking of swaping variables.
This subject was already spoken in the other thread. Just that 
Re: Finding LCM & GCD in C
:) hi !
thanks for your idea. But I only want someone tell me about that algorithm. Swapping is not difficult, but I'm interested in the easier way solving it. Code:
tmp = ts[i]; 
Re: Finding LCM & GCD in C
Hi,
This is Bala. I want to know how to calculate GCD of many numbers, ex 2048 values. Please help. Thanks in advance. 
Re: Finding LCM & GCD in C
Code:
int main() 
Re: Finding LCM & GCD in C
can you please tell me when the loop will actually stop

Re: Finding LCM & GCD in C
correct code is this 100% checked!
for(n=1;n%a != 0  n%b != 0;n++); return n 
Re: Finding LCM & GCD in C
When I searched for lcm c++, this thread had the top two spots, so I figured that I'd join and put this here.
For finding the LCM (Least Common Multiple) of two integers, a and b, in C++, use: for(n=a;n%b != 0;n+=a); return n; For maximum speed, try to set it up so that the larger number is "a" and the smaller one is "b". The reasoning: The least common multiple (LCM) of two numbers, a and b, is the smallest number that is a multiple of "a" AND a multiple of "b". If we already know that the LCM is a multiple of "a", then why not count by "a"? When we count by "a", starting at "a", we are testing every number that is a multiple of "a" already, and only have to see if it is a multiple of "b" too. This code does just that. When "a", the number we're counting by, is larger than 1, the process will be much faster than with those other suggestions, because it's taking bigger steps (at the same speed per step) to get to the same solution. If "a" is a billion, then this will be at least a billion times faster than this: for(n=1;n%a != 0  n%b != 0;n++); return n; (even though both work) I hope this helps. 
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