求GCD和LCM

辗转相除法求最大公因数
对于两个数a&＃xff0c;b来说&＃xff0c;如果a > b&＃xff0c;有 a &＃61; q*b &＃43; r&＃xff0c;则&＃xff08;a&＃xff0c; b&＃xff09;&＃61;&＃xff08;b&＃xff0c; r&＃xff09;&＃xff1b;一直重复下去知道r等于0&＃xff0c;则最后一个不为0的r就是&＃xff08;a&＃xff0c; b&＃xff09;

package gcdAndLCM;
import java.util.Scanner;
//从键盘上输入数据
public class GCD {static int Gcd (int a, int b ) {int r &＃61; 0;if (b > a) {int t &＃61; a;a &＃61; b;b &＃61; t;}else {r &＃61; a % b;while (r !&＃61; 0) {a &＃61; b;b &＃61; r;r &＃61; a % b;}}return b;}public static void main(String[] args) {Scanner scanner &＃61; new Scanner(System.in);System.out.println("输入整数a和b&＃xff1a;");int a &＃61; scanner.nextInt();int b &＃61; scanner.nextInt();//这三行是用来从控制台输入a和b的整数值的//nextLine()输入字符串&＃xff08;换行为间隔&＃xff09;&＃xff1b; nextDouble()输入浮点数 &＃xff1b; next()输入字符串&＃xff08;空格为间隔&＃xff09;scanner.close();//使用玩Scanner之后一定要调用close()函数&＃xff0c;否则会出现警告&＃xff1b;int gcd &＃61; Gcd(a, b);System.out.println("最大公因数是:");System.out.println(gcd);}
}

LCM

求出GCD&＃xff0c;用 a*b / GCD。