Given two positive integers, ask for the maximum number of conventions, for this algorithm, I must know that the most classic way to solve this problem is to use the Ou Sri German algorithm, which is said to be the earliest algorithm in history.
#include
#include
#define
MAX (A, B) (a)> (b)? (a): (b)
#define
MIN (A, B) (a) <(b)? (a): (b)
/ *
INPUT: M, N (m> 0, n> 0)
Output: -1 Error, else the commit divisor of m and n
* /
int
G * c * d (int M, int N)
{
INT A = max (m, n);
INT b = min (m, n);
Int remainder = 1;
IF (m <= 0 || n <= 0)
{
Return -1;
}
While (remainder> 0)
{
Remainder = a% B;
A = B;
B = remainder;
}
Return A;
}
int
Main (int Argc, char * argv [])
{
Int first, second, result
IF (argc! = 3)
{
FPRINTF (stderr, "there" there ");
EXIT (-1);
}
First = ATOI (Argv [1]);
Second = ATOI (Argv [2]);
Result = g * c * d (first, second);
IF (Result! = -1)
{
Printf ("THE COMMON DIVISOR OF% D and% D IS% D / N", First, Second, Result;
}
Else
{
Printf ("M and N Should Be Greater Than 0 / N");
}
Return 0;
}
