Question

有人可以举例说明找到两个以上数字的最大公约数算法吗？

我相信编程语言并不重要。

Answer 1

从第一对开始并获取他们的GCD，然后获取该结果的GCD和下一个数字。显而易见的优化是，如果正在运行的GCD达到1，你可以停止。我正在看这个，看看是否还有其他优化。：）

哦，这很容易并行化，因为操作是可交换的/关联的。

Answer 2

3个数字的GCD可以计算为gcd(a, b, c) = gcd(gcd(a, b), c)。您可以迭代地应用欧几里德算法，扩展欧几里德算法或二进制GCD算法并得到答案。不幸的是，我不知道有任何其他（更聪明的？）找到GCD的方法。

Answer 3

我认识的派对有点晚了，但是使用了Sam Harwell对该算法的描述，这是一个简单的JavaScript实现：

function euclideanAlgorithm(a, b) {
    if(b === 0) {
        return a;
    }
    const remainder = a % b;
    return euclideanAlgorithm(b, remainder)
}

function gcdMultipleNumbers(...args) { //ES6 used here, change as appropriate
  const gcd = args.reduce((memo, next) => {
      return euclideanAlgorithm(memo, next)}
  );

  return gcd;
}

gcdMultipleNumbers(48,16,24,96) //8

Answer 4

在Java中（非最佳）：

public static int GCD(int[] a){
    int j = 0;

    boolean b=true;
    for (int i = 1; i < a.length; i++) {
        if(a[i]!=a[i-1]){
            b=false;
            break;
        }
    }
    if(b)return a[0];
    j=LeastNonZero(a);
    System.out.println(j);
    for (int i = 0; i < a.length; i++) {
        if(a[i]!=j)a[i]=a[i]-j;
    }
    System.out.println(Arrays.toString(a));
    return GCD(a);
}

public static int LeastNonZero(int[] a){
    int b = 0;
    for (int i : a) {
        if(i!=0){
            if(b==0||i<b)b=i;
        }
    }
    return b;
}

Answer 5

我刚刚更新了维基页面。

[https://en.wikipedia.org/wiki/Binary_GCD_algorithm#C.2B.2B_template_class]

这需要任意数量的术语。使用GCD（5,2,30,25,90,12）;

template<typename AType> AType GCD(int nargs, ...)
{
    va_list arglist;
    va_start(arglist, nargs);

    AType *terms = new AType[nargs];

    // put values into an array
    for (int i = 0; i < nargs; i++) 
    {
        terms[i] = va_arg(arglist, AType);
        if (terms[i] < 0)
        {
            va_end(arglist);
            return (AType)0;
        }
    }
    va_end(arglist);

    int shift = 0;
    int numEven = 0;
    int numOdd = 0;
    int smallindex = -1;

    do
    {
        numEven = 0;
        numOdd = 0;
        smallindex = -1;

        // count number of even and odd
        for (int i = 0; i < nargs; i++)
        {
            if (terms[i] == 0)
                continue;

            if (terms[i] & 1)
                numOdd++;
            else
                numEven++;

            if ((smallindex < 0) || terms[i] < terms[smallindex])
            {
                smallindex = i;
            }
        }

        // check for exit
        if (numEven + numOdd == 1)
            continue;

        // If everything in S is even, divide everything in S by 2, and then multiply the final answer by 2 at the end.
        if (numOdd == 0)
        {
            shift++;
            for (int i = 0; i < nargs; i++)
            {
                if (terms[i] == 0)
                    continue;

                terms[i] >>= 1;
            }
        }

        // If some numbers in S  are even and some are odd, divide all the even numbers by 2.
        if (numEven > 0 && numOdd > 0)
        {
            for (int i = 0; i < nargs; i++)
            {
                if (terms[i] == 0)
                    continue;

                if ((terms[i] & 1)  == 0) 
                    terms[i] >>= 1;
            }
        }

        //If every number in S is odd, then choose an arbitrary element of S and call it k.
        //Replace every other element, say n, with | n−k | / 2.
        if (numEven == 0)
        {
            for (int i = 0; i < nargs; i++)
            {
                if (i == smallindex || terms[i] == 0)
                    continue;

                terms[i] = abs(terms[i] - terms[smallindex]) >> 1;
            }
        }

    } while (numEven + numOdd > 1);

    // only one remaining element multiply the final answer by 2s at the end.
    for (int i = 0; i < nargs; i++)
    {
        if (terms[i] == 0)
            continue;

        return terms[i] << shift;
    }
    return 0;
};

Answer 6

对于golang，使用余数

func GetGCD(a, b int) int {
    for b != 0 {
        a, b = b, a%b
    }
    return a
}
func GetGCDFromList(numbers []int) int {
    var gdc = numbers[0]
    for i := 1; i < len(numbers); i++ {
        number := numbers[i]
        gdc  = GetGCD(gdc, number)
    }
    return gdc
}

欧几里德对两个以上数字的最大公约数

6 个答案: