Chapter 2: find the classical solution of the maximum Convention and the least common multiple of a and B, find the conventional solution of the maximum Convention and the least common multiple of a a

// seek a,b Classical solution of maximum Convention and least common multiple of 
int main()
{
    
	long a, b, c, r, a1, b1;
	printf("  Please enter a positive integer  a,b: ");
	scanf("%ld,%ld", &a, &b);
	if (a < b)
	{
    
		c = a;
		a = b;
		b = c;
	}
	a1 = a;
	b1 = b;
	r = a % b;
	while (r != 0) // The implementation of “ division algorithm 
	{
    
		a = b;
		b = r;
		r = a % b;
	}
	printf("(%ld,%ld) = %ld\n", a1, b1, b);

	printf("{%ld,%ld} = %ld\n", a1, b1, a1 * b1 / b);
	return 0;
}

result ：

// seek a,b The conventional solution of the maximum Convention and the least common multiple of 
int main()
{
    
	long a, b, c;
	printf("  Please enter a positive integer  a,b: ");
	scanf("%ld,%ld", &a, &b);
	if (a < b)
	{
    
		c = a;
		a = b;
		b = c;
	}
	for (c = b; c >= 1; c--)
		if (a % c == 0 && b % c == 0) break;  //C Is the trial quotient factor 
	printf("(%ld,%ld) = %ld\n", a, b, c);

	printf("{%ld,%ld} = %ld\n", a, b, a * b / c);
	return 0;
}

result ：

// seek n The maximum Convention and the least common multiple of positive integers 
int main()
{
    
	int k, n;
	long a, b, c, m[100];
	printf(" Please enter the number of positive integers n: ");
	scanf("%d", &n);
	printf(" Please input... In turn %d A positive integer ：", n);
	for (k = 0; k <= n - 1; k++)
	{
    
		printf("\n Please enter the first  %d A positive integer ：", k + 1);
		scanf(" %ld", &m[k]); // Input raw data  
	}
	b = m[0];
	for (k = 1; k <= n - 1; k++) //  Cycle calculation  n - 1  Time 
	{
    
		a = m[k];
		if (a < b)
		{
    
			c = a;
			a = b;    // In exchange for a、b, Make sure a>b
			b = c;
		}
		for (c = b; c >= 1; c--)
			if (a % c == 0 && b % c == 0) break; // Calculate the greatest common divisor 
		if (c == 1) break;   //  If the greatest common divisor is  1, Exit loop  
		b = c;
	}
		printf(" (%ld", m[0]); //  Output the result of the greatest common divisor  
		for (k = 1; k <= n - 1; k++)
			printf(", %ld", m[k]);
		printf(") = %ld\n", c);
		b = m[0];
		for (k = 1; k <= n - 1; k++) // Cycle calculation  n-1  Time  
		{
    
			a = m[k];
			if (a < b)
			{
    
				c = a;
				a = b;
				b = c;
			}
			for (c = a; c <= a * b; c = c + a)
				if (c % b == 0) break;
			b = c;
		}
		printf("{%ld", m[0]);
		for (k = 1; k <= n - 1; k++)
			printf(",%ld", m[k]);
		printf("}=%ld\n", c);
		return 0;
	}

result ：