问题
What would be the easiest way to calculate Greatest Common Divisor and Least Common Multiple on a set of numbers? What math functions can be used to find this information?
回答1:
I've used Euclid's algorithm to find the greatest common divisor of two numbers; it can be iterated to obtain the GCD of a larger set of numbers.
private static long gcd(long a, long b)
{
while (b > 0)
{
long temp = b;
b = a % b; // % is remainder
a = temp;
}
return a;
}
private static long gcd(long[] input)
{
long result = input[0];
for(int i = 1; i < input.length; i++) result = gcd(result, input[i]);
return result;
}
Least common multiple is a little trickier, but probably the best approach is reduction by the GCD, which can be similarly iterated:
private static long lcm(long a, long b)
{
return a * (b / gcd(a, b));
}
private static long gcd(long[] input)
{
long result = input[0];
for(int i = 1; i < input.length; i++) result = lcm(result, input[i]);
return result;
}
回答2:
There is an Euclid's algorithm for GCD,
public int GCF(int a, int b) {
if (b == 0) return a;
else return (GCF (b, a % b));
}
By the way, a and b should be greater or equal 0, and LCM = |ab| / GCF(a, b)
回答3:
There are no build in function for it. You can find the GCD of two numbers using Euclid's algorithm.
For a set of number
GCD(a_1,a_2,a_3,...,a_n) = GCD( GCD(a_1, a_2), a_3, a_4,..., a_n )
Apply it recursively.
Same for LCM:
LCM(a,b) = a * b / GCD(a,b)
LCM(a_1,a_2,a_3,...,a_n) = LCM( LCM(a_1, a_2), a_3, a_4,..., a_n )
回答4:
If you can use Java 8 (and actually want to) you can use lambda expressions to solve this functionally:
private static int gcd(int x, int y) {
return (y == 0) ? x : gcd(y, x % y);
}
public static int gcd(int... numbers) {
return Arrays.stream(numbers).reduce(0, (x, y) -> gcd(x, y));
}
public static int lcm(int... numbers) {
return Arrays.stream(numbers).reduce(1, (x, y) -> x * (y / gcd(x, y)));
}
I oriented myself on Jeffrey Hantin's answer, but
- calculated the gcd functionally
- used the varargs-Syntax for an easier API (I was not sure if the overload would work correctly, but it does on my machine)
- transformed the gcd of the
numbers-Array into functional syntax, which is more compact and IMO easier to read (at least if you are used to functional programming)
This approach is probably slightly slower due to additional function calls, but that probably won't matter at all for the most use cases.
回答5:
int gcf(int a, int b)
{
while (a != b) // while the two numbers are not equal...
{
// ...subtract the smaller one from the larger one
if (a > b) a -= b; // if a is larger than b, subtract b from a
else b -= a; // if b is larger than a, subtract a from b
}
return a; // or return b, a will be equal to b either way
}
int lcm(int a, int b)
{
// the lcm is simply (a * b) divided by the gcf of the two
return (a * b) / gcf(a, b);
}
回答6:
int lcmcal(int i,int y)
{
int n,x,s=1,t=1;
for(n=1;;n++)
{
s=i*n;
for(x=1;t<s;x++)
{
t=y*x;
}
if(s==t)
break;
}
return(s);
}
回答7:
With Java 8, there are more elegant and functional ways to solve this.
LCM:
private static int lcm(int numberOne, int numberTwo) {
final int bigger = Math.max(numberOne, numberTwo);
final int smaller = Math.min(numberOne, numberTwo);
return IntStream.rangeClosed(1,smaller)
.filter(factor -> (factor * bigger) % smaller == 0)
.map(factor -> Math.abs(factor * bigger))
.findFirst()
.getAsInt();
}
GCD:
private static int gcd(int numberOne, int numberTwo) {
return (numberTwo == 0) ? numberOne : gcd(numberTwo, numberOne % numberTwo);
}
Of course if one argument is 0, both methods will not work.
回答8:
for gcd you cad do as below:
String[] ss = new Scanner(System.in).nextLine().split("\\s+");
BigInteger bi,bi2 = null;
bi2 = new BigInteger(ss[1]);
for(int i = 0 ; i<ss.length-1 ; i+=2 )
{
bi = new BigInteger(ss[i]);
bi2 = bi.gcd(bi2);
}
System.out.println(bi2.toString());
回答9:
Basically to find gcd and lcm on a set of numbers you can use below formula,
LCM(a, b) X HCF(a, b) = a * b
Meanwhile in java you can use euclid's algorithm to find gcd and lcm, like this
public static int GCF(int a, int b)
{
if (b == 0)
{
return a;
}
else
{
return (GCF(b, a % b));
}
}
You can refer this resource to find examples on euclid's algorithm.
回答10:
import java.util.Scanner; public class Lcmhcf {
/**
* @param args the command line arguments
*/
public static void main(String[] args) {
// TODO code application logic here
Scanner scan = new Scanner(System.in);
int n1,n2,x,y,lcm,hcf;
System.out.println("Enter any 2 numbers....");
n1=scan.nextInt();
n2=scan.nextInt();
x=n1;
y=n2;
do{
if(n1>n2){
n1=n1-n2;
}
else{
n2=n2-n1;
}
} while(n1!=n2);
hcf=n1;
lcm=x*y/hcf;
System.out.println("HCF IS = "+hcf);
System.out.println("LCM IS = "+lcm);
}
}
//## Heading ##By Rajeev Lochan Sen
回答11:
import java.util.Scanner;
public class Main {
public static void main(String[] args) {
Scanner input = new Scanner(System.in);
int n0 = input.nextInt(); // number of intended input.
int [] MyList = new int [n0];
for (int i = 0; i < n0; i++)
MyList[i] = input.nextInt();
//input values stored in an array
int i = 0;
int count = 0;
int gcd = 1; // Initial gcd is 1
int k = 2; // Possible gcd
while (k <= MyList[i] && k <= MyList[i]) {
if (MyList[i] % k == 0 && MyList[i] % k == 0)
gcd = k; // Update gcd
k++;
count++; //checking array for gcd
}
// int i = 0;
MyList [i] = gcd;
for (int e: MyList) {
System.out.println(e);
}
}
}
回答12:
import java.util.*;
public class lcm {
public static void main(String args[])
{
int lcmresult=1;
System.out.println("Enter the number1: ");
Scanner s=new Scanner(System.in);
int a=s.nextInt();
System.out.println("Enter the number2: ");
int b=s.nextInt();
int max=a>b?a:b;
for(int i=2;i<=max;i++)
{
while(a%i==0||b%i==0)
{
lcmresult=lcmresult*i;
if(a%i==0)
a=a/i;
if(b%i==0)
b=b/i;
if(a==1&&b==1)
break;
}
}
System.out.println("lcm: "+lcmresult);
}
}
回答13:
int lcm = 1;
int y = 0;
boolean flag = false;
for(int i=2;i<=n;i++){
if(lcm%i!=0){
for(int j=i-1;j>1;j--){
if(i%j==0){
flag =true;
y = j;
break;
}
}
if(flag){
lcm = lcm*i/y;
}
else{
lcm = lcm*i;
}
}
flag = false;
}
here, first for loop is for getting every numbers starting from '2'. then if statement check whether the number(i) divides lcm if it does then it skip that no. and if it doesn't then next for loop is for finding a no. which can divides the number(i) if this happens we don't need that no. we only wants its extra factor. so here if the flag is true this means there already had some factors of no. 'i' in lcm. so we divide that factors and multiply the extra factor to lcm. If the number isn't divisible by any of its previous no. then when simply multiply it to the lcm.
来源：https://stackoverflow.com/questions/4201860/how-to-find-gcd-lcm-on-a-set-of-numbers