0
Go4Expert Member
Lets see the performance.

Which is the fastest algorithm?

GCM seems to be at most n
LCM seems to be at most n*m

so, if you do lcm(a,b) := (a*b)/gcm(a,b), you get a n complexity

This is mathematics.
0
Go4Expert Member
Swamp is the subject of another post:
0
Go4Expert Founder
Quote:
Originally Posted by oleber
Swamp is the subject of another post:
Could not get what you meant here.
0
Go4Expert Member
clocking and shabbir are speaking of swaping variables.

0
Ambitious contributor
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];
ts[i] = ts[j];
ts[j] = tmp;```
can't we use tmp ?
0
Newbie Member
Hi,

This is Bala.
I want to know how to calculate GCD of many numbers, ex 2048 values.

0
TechCake
Code:
```int main()
{
int n1,n2,n3,n;
cout<<" Numbers are : ";
cin>>n1>>n2>>n3;
n=FindGcd(FindGcd(n1,n2),n3);
cout<<"GCD of n1,n2,and n3 is "<<n<<endl;
return 0;
}

int FindGcd(int num1, int num2)
{
int temp;
while(num2!=0)
{
temp = num2;
num2 = num1%num2;
num1 = temp;
}
return num1;
}```

Last edited by shabbir; 27Feb2008 at 13:20.. Reason: Code block
0
Contributor
can you please tell me when the loop will actually stop
0
Newbie Member
correct code is this 100% checked!

for(n=1;n%a != 0 || n%b != 0;n++);
return n
0
Newbie Member
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.