是什么方法计算3个数字的GCD

Question

使用三个数字找到GCD（最大公共分频器）的方法是什么？ 以下代码显示了具有2个数字的方法，该数字使用Euclids算法的基本版本（因为输入为正）来计算GCD。

public class GCD {  
 public  static void main(String[] args) {

  int age1 = 10;
  int age2 = 15;

  int multiple1OfGCD = age1;
  int multiple2OfGCD = age2;



  while (multiple1OfGCD != multiple2OfGCD ) {

   if (multiple1OfGCD > multiple2OfGCD) {
    multiple1OfGCD -= multiple2OfGCD;
   }
   else {
    multiple2OfGCD -= multiple1OfGCD;
   }
  }

  System.out.println("The GCD of " + age1 + " and " + age2 + " is " + multiple1OfGCD);


  int noOfPortions1 = age1 / multiple1OfGCD;
  int noOfPortions2 = age2 / multiple1OfGCD;




  System.out.println("So the cake should be divided into "

     + (noOfPortions1 + noOfPortions2));

  System.out.println("The " + age1 + " year old gets " + noOfPortions1

     + " and the " + age2 + " year old gets " + noOfPortions2);

 } 
}

我希望输出如下图所示：

Answer 1

希望它会有所帮助

 public static void main (String[] args)
{
 int a,b,c;
 a=10;
 b=15;
 c=20;
 int d= gcd(a,b,c);
 System.out.println("The GCD of "+a+", "+b+" and "+c+ " is "+d);
 int cake=a/d+b/d+c/d;
 System.out.println("So the cake is divided into "+ cake);
 System.out.println("The "+a+ " Years old get "+a/d );
 System.out.println("The "+b+ " Years old get "+b/d );
 System.out.println("The "+c+ " Years old get "+c/d );
}

public static int gcd(int a, int b, int c){
 return calculateGcd(calculateGcd(a, b), c);
}

public static int calculateGcd(int a, int b) {
    if (a == 0) return b;
    if (b == 0) return a;
    if (a > b) return calculateGcd(b, a % b);
    return calculateGcd(a, b % a);
 }
}