To solve the-CodePudding

#include Int HCF (int int u, v)//the greatest common divisor, {int r; While ((r=u % v)!=0)//and division

CodePudding user response:

Reference:

 # include & lt; Iostream. H> 
#include  
Int gys (int x, int y)//division algorithm common divisor 
{
int a; 
If (x{
A=x; 
X=y; 
Y=a; 
} 
While (x % y!=0) 
{
A=x % y; 
X=y; 
Y=a; 
} 
The return of y; 
} 

Int GBS (int x, int y)//common multiple 
{
Int result=(x * y)/(gys (x, y)); 
return result; 
} 

Int main (int arg c, char * argv []) 
{
Int x, y; 
Cout<& lt;" Please enter the two positive integers: "; 
Cin> x> y; 

Cout<& lt;" The greatest common divisor for: "& lt; Cout<& lt;" LCM is: "& lt; 
system("pause"); 
return 0; 
}

CodePudding user response:

The code and the demand for such online very much, suggest check myself, try to test