Here you can find the source of gcd(int n, int m)
n and m.
| Parameter | Description |
|---|---|
| n | first number |
| m | second number |
public static int gcd(int n, int m)
//package com.java2s; /*//from ww w.ja va2 s . co m * NOTE: This copyright does *not* cover user programs that use HQ * program services by normal system calls through the application * program interfaces provided as part of the Hyperic Plug-in Development * Kit or the Hyperic Client Development Kit - this is merely considered * normal use of the program, and does *not* fall under the heading of * "derived work". * * Copyright (C) [2004, 2005, 2006], Hyperic, Inc. * This file is part of HQ. * * HQ is free software; you can redistribute it and/or modify * it under the terms version 2 of the GNU General Public License as * published by the Free Software Foundation. This program is distributed * in the hope that it will be useful, but WITHOUT ANY WARRANTY; without * even the implied warranty of MERCHANTABILITY or FITNESS FOR A * PARTICULAR PURPOSE. See the GNU General Public License for more * details. * * You should have received a copy of the GNU General Public License * along with this program; if not, write to the Free Software * Foundation, Inc., 59 Temple Place, Suite 330, Boston, MA 02111-1307 * USA. */ public class Main { /** * Find the greatest common divisor of both <code>n</code> and * <code>m</code>. * * @param n first number * @param m second number * @return the GCD of n and m, or 1 if both numbers are 0 */ public static int gcd(int n, int m) { n = Math.abs(n); m = Math.abs(m); // avoid infinite recursion if (n == 0 && m == 0) { return 1; } if (n == m && n >= 1) { return n; } return (m < n) ? gcd(n - m, n) : gcd(n, m - n); } }