Java lcm lcm(long u, long v)

Description

lcm

License

Open Source License

Declaration

public static long lcm(long u, long v) 

Method Source Code

//package com.java2s;
/**//  ww w  .  j a v a  2s.  c  o  m
 * $Id$
 *
 * Copyright (c) 2009 Thomas Beckmann 
 *
 * Permission is hereby granted, free of charge, to any person obtaining a copy
 * of this software and associated documentation files (the "Software"), to deal
 * in the Software without restriction, including without limitation the rights
 * to use, copy, modify, merge, publish, distribute, sublicense, and/or sell
 * copies of the Software, and to permit persons to whom the Software is
 * furnished to do so, subject to the following conditions:
 *
 * The above copyright notice and this permission notice shall be included in
 * all copies or substantial portions of the Software.
 *
 * THE SOFTWARE IS PROVIDED "AS IS", WITHOUT WARRANTY OF ANY KIND, EXPRESS OR
 * IMPLIED, INCLUDING BUT NOT LIMITED TO THE WARRANTIES OF MERCHANTABILITY,
 * FITNESS FOR A PARTICULAR PURPOSE AND NONINFRINGEMENT. IN NO EVENT SHALL THE
 * AUTHORS OR COPYRIGHT HOLDERS BE LIABLE FOR ANY CLAIM, DAMAGES OR OTHER
 * LIABILITY, WHETHER IN AN ACTION OF CONTRACT, TORT OR OTHERWISE, ARISING FROM,
 * OUT OF OR IN CONNECTION WITH THE SOFTWARE OR THE USE OR OTHER DEALINGS IN THE
 * SOFTWARE.
 */

public class Main {
    public static long lcm(long u, long v) {

        return (u / gcd(u, v)) * v;
    }

    public static int gcd(int u, int v) {

        /* GCD(0,v) := v, GCD(u,0) := u */
        if (u == 0 || v == 0)
            return u | v;

        int shift;

        /* Let shift := lg K, where K is the greatest power of 2
        dividing both u and v. */
        for (shift = 0; ((u | v) & 1) == 0; ++shift) {
            u >>= 1;
            v >>= 1;
        }

        while ((u & 1) == 0)
            u >>= 1;

        /* From here on, u is always odd. */
        do {
            while ((v & 1) == 0) /* Loop X */
                v >>= 1;

            /* Now u and v are both odd, so diff(u, v) is even.
               Let u = min(u, v), v = diff(u, v)/2. */
            if (u < v) {
                v -= u;
            } else {
                int diff = u - v;
                u = v;
                v = diff;
            }
            v >>= 1;
        } while (v != 0);

        return u << shift;
    }

    public static long gcd(long u, long v) {

        /* GCD(0,v) := v, GCD(u,0) := u */
        if (u == 0 || v == 0)
            return u | v;

        long shift;

        /* Let shift := lg K, where K is the greatest power of 2
        dividing both u and v. */
        for (shift = 0; ((u | v) & 1) == 0; ++shift) {
            u >>= 1;
            v >>= 1;
        }

        while ((u & 1) == 0)
            u >>= 1;

        /* From here on, u is always odd. */
        do {
            while ((v & 1) == 0) /* Loop X */
                v >>= 1;

            /* Now u and v are both odd, so diff(u, v) is even.
               Let u = min(u, v), v = diff(u, v)/2. */
            if (u < v) {
                v -= u;
            } else {
                long diff = u - v;
                u = v;
                v = diff;
            }
            v >>= 1;
        } while (v != 0);

        return u << shift;
    }
}

Related

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2. lcm(int x1, int x2)
3. lcm(long a, long b)
4. lcm(long a, long b)
5. lcm(long a, long b)
6. lcm(long x, long y)
7. lcm(long[] a)
8. LCM1(int A, int B)
9. lcmPositive(int... args)