On the OI in number theory

  The way it is, I'm a new sprout, then sprout new beginner number theory. qvq

  This article is about the difficulty of gcd ~ Mobius inversion, maybe I'll write a little combination count, the inclusion-exclusion principle, knowledge of linear algebra, of course, I guess I will not, because the cuckoo.

  Taking mathematical proof based, code is easy to understand, all the operations of computer algorithms prevail. qvq ( other people's code the wind was not cancer )

  GCD&LCM

  That is the greatest common divisor and least common multiple, presented here Euclid seek gcd.

  Set (a, b) represents gcd (a, b).

  Prove that: gcd (a, b) = gcd (b, a% b)

  prove:

  Set a = k1 * c, b = k2 * c and (k1, k2) = 1.

  Then (a, b) = c.

  Provided k3 = a / b (according to computer algorithms, rounded down)

  Provisions a% b = a-k3 * b.

  There is a% b = a-k3 * b = k1 * c-k3 * k2 * c = (k1-k2 * k3) * c.

  若（k2,k1-k2*k3）=1,(b,a%b)=c,则（a,b）=（b,b%a）.

  If (k2, k1-k2 * k3) ≠ 1, provided k2 = m1 * d, k1-k2 * k3 = m2 * d.

  Then (a, b) = (k3 * m1 * d * c + m2 * d * c, m1 * d * c) = dc ≠ c, the assumption does not hold.

  In summary, (a, b) = (b, a% b).

 QED.

In particular, when b = 0 when, (a, b) = a.

LCM method of finding:. Lcm (a, b) = a * b / gcd (a, b) is not demonstrated.

Code:

 

#include<cstdio>
int a,b;
int gcd(int a,int b)
{
    return (b==0)?a:gcd(b,a%b);
}
int lcm(int a,int b)
{
    return a*b/gcd(a,b);
}
int main()
{
    scanf("%d%d",&a,&b);
    printf("GCD=%d LCM=%d\n",gcd(a,b),lcm(a,b));
    return 0;
}

View Code