java 如何在单个方法中找到三个数字的 GCD

How to find the GCD of three numbers within a single method

提问by tech_geek23

I've got to ensure that the GCD between 3 numbers is no greater than 1.

我必须确保 3 个数字之间的 GCD 不大于 1。

Here's the code I have so far for the method:

这是我迄今为止为该方法编写的代码：

private int greatestCommonFactor(int a, int b, int c)
{
    for(int n = 0; n <= number; n++)
    {
        if()
    }

    return 1;
}

the return 1was already there when I started working on the lab. How can I make sure that the GCD is no more than 1? And return all three integers?

return 1当我开始在实验室工作时，它已经在那里了。如何确保 GCD 不超过 1？并返回所有三个整数？

Here's the remainder of the code if it helps in figuring out what needs to be done:

下面是代码的其余部分，如果它有助于弄清楚需要做什么：

import static java.lang.System.*;

public class Triples
{
 private int number;

public Triples()
{
    this(0);
}

public Triples(int num)
{
    number = num;
}

public void setNum(int num)
{
    number = num;
}

private int greatestCommonFactor(int a, int b, int c)
{
    for(int n = 0; n <= number; n++)
    {
        if()
    }

    return 1;
}

public String toString()
{
    String output="";
    int max = number;
    for(a = 1; a <= max; a++)
    {
        for(b = a +1; b <= max; b++)
        {
            for(c = b + 1; c <= max; c++)
            {
                if(Math.pow(a, 2)+ Math.pow(b, 2)== Math.pow(c, 2))
                {
                    if((a%2==1 && b%2==0)|| (a%2==0 && b%2==1))
                }
            }
        }
    }


    return output+"\n";
}
}

UPDATE

更新

Here is my new coding for the same lab:

这是我为同一个实验室编写的新代码：

import static java.lang.System.*;

public class Triples
{
 private int number;

public Triples()
{
    this(0);
}

public Triples(int num)
{
    number = num;
}

public void setNum(int num)
{
    number = num;
}

private int greatestCommonFactor(int a, int b, int c)
{
    for(int n = 0; n <= number; n++)
    {
    int max = number;
    for(a = 1; a <= max; a++)
    {
        a = n;
        for(b = a +1; b <= max; b++)
        {
            b =n;
            for(c = b + 1; c <= max; c++)
            {
                c = n;
                if(Math.pow(a, 2)+ Math.pow(b, 2)== Math.pow(c, 2))
                {
                    if((a%2==1 && b%2==0)|| (a%2==0 && b%2==1))
                    {
                        if(a%2<=1 && b%2<=1 && c%2<=1)
                        {
                            return 1;
                        }
                    }
                }
            }
        }
    }
    }

    return 1;
}

public String toString()
{
    String output="";
    output = greatestCommonFactor(a, b, c);


    return output+"\n";
}
}

回答by Ted Hopp

You can use Euclid's algorithmto calculate the GCD of aand b. Call the result d. Then the GCD of a, b, and cis the GCD of cand d; for that, you can use Euclid's algorithm again.

你可以用欧几里德的算法来计算的GCDa和b。调用结果d。然后的GCD a，b和c是的GCDc和d; 为此，您可以再次使用 Euclid 算法。

回答by sampson-chen

Here's a brute-force way if you don't care about efficiency:

如果您不关心效率，这是一种蛮力方法：

private int greatestCommonFactor(int a, int b, int c)
{
    limit = Math.min(a, b);
    limit = Math.min(limit, c);
    for(int n = limit; n >= 2; n--)
    {
        if ( (a % n == 0) && (b % n == 0) && (c % n == 0) ) {
            return n;
        }
    }

    return 1;
}

Explanation:

解释：

• You can save some work by only checking up to the minimum of (a, b, c). Any number greater than that definitely won't be a GCD of all 3.
• You need to start your loop at n = limitinstead of n = 0and count backwards.
• As soon as we come across a number that produces zero remainder for (a, b, c), that must be the GCD.
• If nothing is found within the loop, GCD defaults to 1.

• 您可以通过最多只检查(a, b, c).任何大于这个的数字肯定不会是所有 3 的 GCD。
• 您需要在n = limit而不是开始循环n = 0并倒数。
• 一旦我们遇到一个对 产生零余数的数字(a, b, c)，那一定是 GCD。
• 如果在循环中没有找到任何东西，GCD 默认为 1。