Find the greatest common divisor and least common multiple of two numbers

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greatest common divisor

The solution of violence

Two, tossing and dividing method

Least common multiple

The solution of violence

Two, tossing and dividing method

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greatest common divisor

The solution of violence

#define _CRT_SECURE_NO_WARNINGS 1
#include <stdio.h>

int main()
{
	int num1 = 0;
	int num2 = 0;
	scanf("%d%d", &num1, &num2);
	int min = (num1 < num2 ? num1 : num2);
	while (1)
	{
		if (num1 % min == 0 && num2 % min == 0)
		{
			break;
		}
		min--;
	}
	printf("最大公约数:%d\n", min);
	return 0;
}

Two, tossing and dividing method

#define _CRT_SECURE_NO_WARNINGS 1
#include <stdio.h>

int main()
{
	int num1 = 0;
	int num2 = 0;
	scanf("%d%d", &num1, &num2);
	int k = 0;
	while (k = num1 % num2)
	{
		num1 = num2;
		num2 = k;
	}
	printf("最大公约数:%d\n", num2);
	return 0;
}

Least common multiple

The solution of violence

int main()
{
	int num1 = 0;
	int num2 = 0;
	scanf("%d%d", &num1, &num2);
	int max = num1 > num2 ? num1 : num2;
	int i = 0;
	for (i = max;; i++)
	{
		if (i % num1 == 0 && i % num2 == 0)
			break;
	}
	printf("最小公倍数:%d\n", i);

	return 0;
}

Two, tossing and dividing method

First, find the greatest common divisor using the method of rolling division , and then divide the product of the two numbers by the greatest common divisor to obtain the least common multiple.

#define _CRT_SECURE_NO_WARNINGS 1
#include <stdio.h>

int main()
{
	int num1 = 0;
	int num2 = 0;
	scanf("%d%d", &num1, &num2);
	int m = num1;
	int n = num2;

	int k = 0;
	while (k = num1 % num2)
	{
		num1 = num2;
		num2 = k;
	}
	printf("最小公倍数:%d\n", m * n / num2);
	return 0;
}

Note : Two numbers a, b, the greatest common divisor and the least common multiple are m and n, then a*b=m*n.