Finding LCM & GCD in C

Discussion in 'C' started by pradeep, Oct 5, 2005.

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1. oleberNew Member

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Lets see the performance.

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

This is mathematics.

2. oleberNew Member

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4. oleberNew Member

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clocking and shabbir are speaking of swaping variables.

5. clockingNew Member

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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 ?

6. BalaselviNew Member

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Hi,

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

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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 a moderator: Feb 27, 2008
8. aisha.ansari84New Member

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can you please tell me when the loop will actually stop

9. zeemeNew Member

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correct code is this 100% checked!

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

10. vanishing_staplerNew Member

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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.

11. jackspaNew Member

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For lcm try this one:
int lcm(int a,int b)
{
int n;
if(a<b)
{
n=a;
a=b;
b=n;
}
for(n=a;n%b!=0;n+=a)
return n;
}:happy:

12. xpi0t0sMentor

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After 8 posts I would have thought you would know about code blocks by now. USE THEM PLEASE.

13. jackspaNew Member

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For lcm try this one:
Code:
```int lcm(int a;int b)
{
int n;
if(a<b)
{
n=a;
a=b;
b=n;
}
for(n=a;n%b!=0;n+=a)
return n;
} ```
:happy:

14. jackspaNew Member

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i had written that function using blocks.i don't know what happened when i posted it.

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16. jackspaNew Member

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thank you.

17. ZaiberNew Member

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Actually, I used the code on the first page to correct my code 'til I got it right. Needed this for a take home exam for C class.

Thanks, and in return I'll give you my code which might help someone.

Code:
```int gcd( int x, int y) {
if ( y == 0 ) {
return x;
}
else if (x%y == 0) {
return y;
}
else {
return gcd(y,x%y);
}
}```

18. COKEDUDENew Member

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Wouldn't this code also cover both cases of GCD whether your first variable is bigger or your second variable is bigger?

Code:
```void gcd(int x, int y)
{
if (x > y)
{
int c;
while(1)
{
c = x % y;
return y;
x = y;
y = c;
}
if (y > x)
{
int d;
while(1)
{
d = x % y;
return x;
y = x;
x = d;
}

}
}```

19. ZaiberNew Member

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Mine worked for both instances ok.

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