(column 23) typical C language problem: find the minimum common multiple and maximum common divisor of two numbers. (two solutions)

give the result as follows ：

 

---

The first one is ：

#include<stdio.h>
int main()
{
	int min(int a,int b); // Function find minimum 
	int max(int c,int d); // Function to find the maximum 

	int x,y,i,j=0,sum,num,v;
	printf(" Please enter two numbers ：");
	scanf("%d %d",&x,&y);

    for(i=2;i<=min(x,y);i++) 
    {
    	if(x%i==0&&y%i==0) // Whether there is a number divided by two numbers at the same time before the minimum value of two numbers 
    	{
    		if(i>j) 
			{
    			j=i; // Find the greatest common divisor 
			}
		}
	}
	sum=x*y; // The least common multiple is the product of two numbers 

	for(v=max(x,y);v<=x*y;v++)
	{
		if(v%x==0&&v%y==0) // Find the number divisible 
        {
			if(v<sum)
            {
				sum=v; // Find the least common multiple 
			}
		}
	}
	printf(" greatest common divisor ：%d",j);
	printf("\n");	
	printf(" Minimum common multiple ：%d",sum);
	return 0;	
} 

int min(int a,int b)
{
	return (a<b?a:b);
}
int max(int c,int d)
{
	return (c>d?c:d);
}

---

The second kind ：

#include<stdio.h>
int main()
{
    int a,b,t,r,n;
    printf(" Please enter two numbers ：\n");
    scanf("%d %d",&a,&b);

    if(a<b) // Sort the two numbers from large to small 
    {
        t=b;
        b=a;
        a=t;
    }

    r=a%b; // The least common divisor of two numbers 
    n=a*b; // The greatest common multiple of two numbers 

    while(r!=0)
    {
        a=b;
        b=r;
        r=a%b;
    }
    printf(" The greatest common divisor of these two numbers is %d, The minimum common multiple is %d\n",b,n/b);
    
    return 0;
}