歐幾裡德算法:gcd(a,b) == gcd(a,(b mod a)):其中b>=a,我們首先證明下這個:
令c=gcd(a,b),則存在整數k1,k2使 k1*c=a、k2*c=b;
令r=b mod a,則存在k3使得k3*a+r=b,将上面的a、b分别代入可得到:r=(k2-k1*k3)*c,即r(b mod a)是a、b最大公約數c的倍數。相應的代碼為:
package com.daelly.algorithm.hcf;
public class Main {
public static void main(String[] args) {
System.out.println(gcd(6, 4));
System.out.println(gcd(6, 12));
System.out.println(lcm(10, 8));
}
/**
* 歐幾裡德算法:
* gcd(a, b) == gcd((b % a), a)
* @param a
* @param b
* @return
*/
public static int gcd(int a, int b) {
if(a > b) {//交換
a = a + b;
b = a - b;//after this b become a
a = a - b;
}
int r = 0;
while (a != 0) {
r = b % a;
b = a;
a = r;
}
return b;
}
/**
* 最小公倍數,基于最大公約數來求
* @param a
* @param b
* @return
*/
public static int lcm(int a, int b) {
return (a * b) / gcd(a, b);
}
}