euclid algorithm(greast common divisor)

#include <iostream>using namespace std;void swap(int &a, int &b){    int tmp;    tmp = a;    a = b;    b = tmp;}int euclid_fun(int m ,int n){    if (n > m)        swap(m, n);    int r = m % n;    while(r != 0)    {        m = n;        n = r;        r = m % n;    }    return n;}int main(){    int m = 24;    int n = 30;    int gcd = euclid_fun(m,n);    cout << " the greatest common factor is : " << gcd << endl;    return 0;}

#include <iostream>using namespace std;void swap(int &a, int &b){    int t;    t = a;    a = b;    b = t;}int fuclid_fun(int &a, int &b){    if (b > a)        swap(a,b);    a = a % b;    if (a == 0){        return b;    }else {        b = b % a;        if ( b == 0)        {            return a;        }        else{            fuclid_fun( a, b);        }    }}int main(){    int a = 2166;    int b = 6099;    int gcd = fuclid_fun (a ,b);    cout << gcd << endl;    return 0;}

Proof of validity[1]

The validity of the Euclidean algorithm can be proven by a two-step argument.^[16] In the first step, the final nonzero remainder r_N−1 is shown to divide both a and b. Since it is a common divisor, it must be less than or equal to the greatest common divisor g. In the second step, it is shown that any common divisor of a and b, including g, must divide r_N−1; therefore, g must be less than or equal to r_N−1. These two conclusions are inconsistent unless r_N−1 = g.

To demonstrate that r_N−1 divides both a and b (the first step), r_N−1 divides its predecessor r_N−2

r_N−2 = q_N r_N−1

since the final remainder r_N is zero. r_N−1 also divides its next predecessor r_N−3

r_N−3 = q_N−1 r_N−2 + r_N−1

because it divides both terms on the right-hand side of the equation. Iterating the same argument, r_N−1 divides all the preceding remainders, including a and b. None of the preceding remainders r_N−2, r_N−3, etc. divide a and b, since they leave a remainder. Since r_N−1 is a common divisor of a and b, r_N−1 ≤ g.

In the second step, any natural number c that divides both a and b (in other words, any common divisor of a and b) divides the remainders r_k. By definition, a and b can be written as multiples of c: a = mc and b = nc, where m and n are natural numbers. Therefore, c divides the initial remainder r₀, since r₀ = a − q₀b = mc − q₀nc = (m − q₀n)c. An analogous argument shows that c also divides the subsequent remainders r₁, r₂, etc. Therefore, the greatest common divisor g must divide r_N−1, which implies that g ≤ r_N−1. Since the first part of the argument showed the reverse (r_N−1 ≤ g), it follows that g = r_N−1. Thus, g is the greatest common divisor of all the succeeding pairs:^[17][18]

g = gcd(a, b) = gcd(b, r₀) = gcd(r₀, r₁) = … = gcd(r_N−2, r_N−1) = r_N−1.

example

a = 1071 and b = 462

[1]. http://en.wikipedia.org/wiki/Euclidean_algorithm