The method of GCD and LCM of two numbers in C language _c language

GCD of two positive integers

Train of thought: This is a very basic problem, the most common is two methods, Euclidean method and subtraction. The formula is f (x, y) = f (y, x%y), f (x, y) = f (y, XY) (x >=y > 0). It is not difficult to write an algorithm according to the formula, which is not given here. Here is the "beauty of programming" algorithm, mainly to reduce the number of iterations.
For x and y, if y = k * y1, x= k * X1, then f (x, y) = k * F (x1, y1). In addition, if x = p * x1, assuming P is prime and y% p!= 0, then f (x, y) = f (p * x1, y) = f (x1, y). Take P = 2.

Reference code:

function function: Find GCD 
//function parameter: X,y is two number 
//return value:  gcd 
int gcd_solution1 (int x, int y) 
{ 
  if (y = = 0) 
    return x; 
  else if (x < y) return 
    gcd_solution1 (y, x); 
  else 
  { 
    if (x&1)//x is odd 
    { 
      if (y&1)//y is odd return 
        gcd_solution1 (y, xy); 
      Else  //y is an even return 
        gcd_solution1 (x, y>>1) 
    ; 
    Else//x is an even 
    { 
      if (y&1)//y is an odd return 
        Gcd_solution1 (x>>1, y); 
      Else  //y is an even return 
        Gcd_solution1 (x>>1, y>>1) << 1 
    } 
  } 
} 

Ask LCM:
The most commonly used is the Euclidean method, which has two integers a and B:
①a%b to the remainder C
② if c=0, then B is two number of GCD
③ if c≠0, then a=b,b=c, then go back to execute ①

The following are not recursive versions:

int gcd_solution2 (int x, int y) 
{ 
  int result = 1; 
  while (y) 
  { 
    int t = x; 
    if (x&1) 
    { 
      if (y&1) 
      { 
        x = y; 
        y = t% y; 
      } 
      else 
        y >>= 1; 
    } 
    else 
    { 
      if (y&1) 
        x >>= 1; 
      else 
      { 
        x >>= 1; 
        Y >>= 1; 
        Result <<= 1; 
  }} return result * x; 
}