# Calculate the maximum common divisor gcd (A, B) of two numbers & proves Euclidean Algorithm

Source: Internet
Author: User

Calculate the maximum common divisor of two numbers A and B. We can think of enumerating each positive integer from [1, min (A, B:

`#include<iostream>using namespace std;int gcd(int a,int b){    int ans=1;    for(int i=2;i<=min(a,b);++i)    {        if(a%i==0 && b%i==0)            ans=i;    }    return ans;}int main(){    int a,b;    cin>>a>>b;    cout<<gcd(a,b);    return 0;}`

However, when a and B are large, this algorithm is not fast enough. There are faster and more elegant algorithms.

• First, a theorem is given:

Gcd (a, B) = gcd (B, A-B) (a> = B)

Proof:

Set gcd (a, B) = M (M> = 1)

Then a % m = 0, B % m = 0

(A % m-B % m) % m = 0

(A-B) % m = 0

Because a % m = 0, B % m = 0, (a-B) % m = 0, gcd (a, B) = m

Therefore, gcd (a, B, a-B) = m;

The following example shows that gcd (B, a-B) = M, and then gcd (a, B) = gcd (B, A-B) is obtained ).

Set c = A-B

Because gcd (a, B, c) = m;

So gcd (B, c )! A sufficient condition for = m is that there is a number D (D> = 1) so that (B/M) % d = 0 and (C/m) % d = 0 and (A/m) % d! = 0.

The following uses the reverse verification method:

D

(C/M) % d = 0

(A-B)/m) % d = 0

(A/m)-(B/M) % d = 0

(A/m) % d-(B/M) % d = 0

Known (B/M) % d = 0, substituted (A/m) % d = 0

Also known (A/m) % d! = 0, so (A/m) % d results belong to (0, d)

X is (0, D), and X % d cannot be equal to 0. Therefore, this is a conflict.

So there is no such d

Therefore, gcd (B, a-B) = gcd (B, c) = m

Gcd (a, B) = gcd (B, A-B) (a> = B) the theorem has been proved.

Therefore, we can use this algorithm for calculation, where gcd (A, 0) =:

`#include<iostream>using namespace std;int gcd(int a,int b){    if(b==0)        return a;    if(a<b)        swap(a,b);    return gcd(b,a-b);}int main(){    int a,b;    cin>>a>>b;    if(a<b)        swap(a,b);    cout<<gcd(a,b);            return 0;}`

Of course, when the data size is large, the stack may overflow, so change it to non-recursive mode.

Can it be faster? (Thank you for your proof)

• Then the second theorem is given:

Gcd (a, B) = gcd (B, A-K * B) where K is 0, 1, 2, 3, 4... and a> = K * B

This theorem proves the same as above.

• The following theorem can be obtained through simplification:

Gcd (a, B) = gcd (B, A % B)

This is the moving phase division (Euclidean Algorithm ).

`#include<iostream>using namespace std;int gcd(int a,int b){    if(b==0)        return a;    return gcd(b,a%b);}int main(){    int a,b;    cin>>a>>b;    cout<<gcd(a,b);            return 0;}`

Calculate the maximum common divisor gcd (A, B) of two numbers & proves Euclidean Algorithm

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