Main.java Source code

Introduction

Here is the source code for Main.java

Source

import java.io.File;

/* 
 * Licensed to the Apache Software Foundation (ASF) under one or more
 *  contributor license agreements.  See the NOTICE file distributed with
 *  this work for additional information regarding copyright ownership.
 *  The ASF licenses this file to You under the Apache License, Version 2.0
 *  (the "License"); you may not use this file except in compliance with
 *  the License.  You may obtain a copy of the License at
 *
 *      http://www.apache.org/licenses/LICENSE-2.0
 *
 *  Unless required by applicable law or agreed to in writing, software
 *  distributed under the License is distributed on an "AS IS" BASIS,
 *  WITHOUT WARRANTIES OR CONDITIONS OF ANY KIND, either express or implied.
 *  See the License for the specific language governing permissions and
 *  limitations under the License.
 *
 *
 */
public class Main {

    /**
     * Returns the least common multiple between two integer values.
     * 
     * @param a the first integer value.
     * @param b the second integer value.
     * @return the least common multiple between a and b.
     * @throws ArithmeticException if the lcm is too large to store as an int
     * @since 1.1
     */
    public static int lcm(int a, int b) {
        return Math.abs(mulAndCheck(a / gcd(a, b), b));
    }

    /**
     * Multiply two integers, checking for overflow.
     * 
     * @param x a factor
     * @param y a factor
     * @return the product <code>x*y</code>
     * @throws ArithmeticException if the result can not be represented as an
     *         int
     * @since 1.1
     */
    public static int mulAndCheck(int x, int y) {
        long m = ((long) x) * ((long) y);
        if (m < Integer.MIN_VALUE || m > Integer.MAX_VALUE) {
            throw new ArithmeticException("overflow: mul");
        }
        return (int) m;
    }

    /**
     * <p>
     * Gets the greatest common divisor of the absolute value of two numbers,
     * using the "binary gcd" method which avoids division and modulo
     * operations. See Knuth 4.5.2 algorithm B. This algorithm is due to Josef
     * Stein (1961).
     * </p>
     * 
     * @param u a non-zero number
     * @param v a non-zero number
     * @return the greatest common divisor, never zero
     * @since 1.1
     */
    public static int gcd(int u, int v) {
        if (u * v == 0) {
            return (Math.abs(u) + Math.abs(v));
        }
        // keep u and v negative, as negative integers range down to
        // -2^31, while positive numbers can only be as large as 2^31-1
        // (i.e. we can't necessarily negate a negative number without
        // overflow)
        /* assert u!=0 && v!=0; */
        if (u > 0) {
            u = -u;
        } // make u negative
        if (v > 0) {
            v = -v;
        } // make v negative
          // B1. [Find power of 2]
        int k = 0;
        while ((u & 1) == 0 && (v & 1) == 0 && k < 31) { // while u and v are
                                                         // both even...
            u /= 2;
            v /= 2;
            k++; // cast out twos.
        }
        if (k == 31) {
            throw new ArithmeticException("overflow: gcd is 2^31");
        }
        // B2. Initialize: u and v have been divided by 2^k and at least
        // one is odd.
        int t = ((u & 1) == 1) ? v : -(u / 2)/* B3 */;
        // t negative: u was odd, v may be even (t replaces v)
        // t positive: u was even, v is odd (t replaces u)
        do {
            /* assert u<0 && v<0; */
            // B4/B3: cast out twos from t.
            while ((t & 1) == 0) { // while t is even..
                t /= 2; // cast out twos
            }
            // B5 [reset max(u,v)]
            if (t > 0) {
                u = -t;
            } else {
                v = t;
            }
            // B6/B3. at this point both u and v should be odd.
            t = (v - u) / 2;
            // |u| larger: t positive (replace u)
            // |v| larger: t negative (replace v)
        } while (t != 0);
        return -u * (1 << k); // gcd is u*2^k
    }
}