Description
Returns the least common multiple between two integer values.
License
Apache License
Parameter
| Parameter | Description |
|---|
| a | the first integer value. |
| b | the second integer value. |
Exception
| Parameter | Description |
|---|
| ArithmeticException | if the lcm is too large to store as an int |
Return
the least common multiple between a and b.
Declaration
public static int lcm(int a, int b)
Method Source Code
//package com.java2s;
/*//from w w w .j a v a2 s . c o m
* 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;
}
/**
* Multiply two long integers, checking for overflow.
*
* @param a first value
* @param b second value
* @return the product <code>a * b</code>
* @throws ArithmeticException if the result can not be represented as an
* long
* @since 1.2
*/
public static long mulAndCheck(long a, long b) {
long ret;
String msg = "overflow: multiply";
if (a > b) {
// use symmetry to reduce boundry cases
ret = mulAndCheck(b, a);
} else {
if (a < 0) {
if (b < 0) {
// check for positive overflow with negative a, negative b
if (a >= Long.MAX_VALUE / b) {
ret = a * b;
} else {
throw new ArithmeticException(msg);
}
} else if (b > 0) {
// check for negative overflow with negative a, positive b
if (Long.MIN_VALUE / b <= a) {
ret = a * b;
} else {
throw new ArithmeticException(msg);
}
} else {
// assert b == 0
ret = 0;
}
} else if (a > 0) {
// assert a > 0
// assert b > 0
// check for positive overflow with positive a, positive b
if (a <= Long.MAX_VALUE / b) {
ret = a * b;
} else {
throw new ArithmeticException(msg);
}
} else {
// assert a == 0
ret = 0;
}
}
return ret;
}
/**
* <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
}
}
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