[6] Program
Title: Enter two positive integers m and n, and seeking the least common multiple of the greatest common divisor.
In the cycle, as long as the divisor is not equal to 0, the smaller number with a larger number by the smaller number as a next cycle of large numbers,
the remainder obtained as the next cycle of the smaller number, so loop until the smaller number is 0,
return a large number, this number is the greatest common divisor of the least common multiple of the product divided by the greatest common divisor of two numbers.
package case50;
import java.util.Scanner;
/**
*
* 【程序6】
* 题目：输入两个正整数m和n，求其最大公约数和最小公倍数。
* 在循环中，只要除数不等于0，用较大数除以较小的数，将小的一个数作为下一轮循环的大数，
* 取得的余数作为下一轮循环的较小的数，如此循环直到较小的数的值为0，
* 返回较大的数，此数即为最大公约数，最小公倍数为两数之积除以最大公约数。
* @author 眼睫毛能扫地
*
*/
//计算最大公约数
class CalGCD {
public int CalGCD(int x, int y) {
int t = 0;
if (x > y) {
t = x;
x = y;
y = t;
}
while (x != 0) {
if (x == y)
return y;
else {
int k = y % x;
y = x;
x = k;
}
}
return y;
}
}
public class Case06 {
public static void main(String[] args) {
int num1 = 0, num2 = 0, gcd = 0, lcm = 0;// gcd 最大公约数 lcm 最小公倍数
Scanner input = new Scanner(System.in);
System.out.println("请输入一个数字");
num1 = input.nextInt();
System.out.println("请再输入一个数字");
num2 = input.nextInt();
input.close();
CalGCD GCD = new CalGCD();
gcd = GCD.CalGCD(num1, num2);
lcm = num1 * num2 / gcd;
System.out.println("最大公约数：" + gcd + " 最小公倍数： " + lcm);
}
}