Returns the maximum common number of N numbers.

There are many methods to calculate the maximum common divisor of two numbers, as shown in the following recursion method:

 Int  Gcd (  Int  N,  Int  M)
{
  If  (N  <  M)
{
N  =  M  +  N;
M  =  N -  M;
N  =  N  -  M;
}
  If  (M  =  0  )  Return  N;
  Return  Gcd (m, n  %  M );
} 

Below I use an array to write a maxtimedest that calculates nCode:

 Int  Max_common_divisor (  Int  Several [],  Int  N)
{
  Int  A  =  Several [  0  ];
  Int  B  = Several [  1  ];
  Int  C  =  Gcd (A, B );
  Int  I;
  For  (I  =  2  ; I  <  N; I  ++ )
{
C  =  Gcd (C, several [I]);
}
  Return  C;
} 

Write a main function to test it:

 # Include  <  Stdio. h  >  
# Include  <  Stdlib. h  >  

  Int Gcd (  Int  A,  Int  B );  //  Returns the maximum common divisor of two numbers.  
  Int  Max_common_divisor (  Int  Several [],  Int  N );  //  Returns the maximum common number of N numbers.  
 
  Int  Main ()
{
  Int  A [  4  ];
  Int  X, Y;
A [  0  ]  =  147  ;
A [  1  ]  =  210 ;
A [  2  ]  =  315  ;
A [  3  ]  =  84  ;
Printf (  "  The maximum number of common appointments is % d \ n.  "  , Max_common_divisor (,  4  ));
 Return     0  ;
} 

The test results are as follows: