两个以上整数集合的最大公约数

这里有一个使用LINQ和GCD方法的代码示例,来自您链接的问题。它使用的是其他答案中描述的理论算法。。。 GCD(a, b, c) = GCD(GCD(a, b), c)

static int GCD(int[] numbers)
{
    return numbers.Aggregate(GCD);
}

static int GCD(int a, int b)
{
    return b == 0 ? a : GCD(b, a % b);
}

您可以使用GCD的这个公共属性:

GCD(a, b, c) = GCD(a, GCD(b, c)) = GCD(GCD(a, b), c) = GCD(GCD(a, c), b)

GCD(a, b) 已经定义,很容易概括:

public class Program
{
    static void Main()
    {
        Console.WriteLine(GCD(new[] { 10, 15, 30, 45 }));
    }

    static int GCD(int a, int b)
    {
        return b == 0 ? a : GCD(b, a % b);
    }

    static int GCD(int[] integerSet)
    {
        return integerSet.Aggregate(GCD);
    }    
}

这是C版本。

  public static int Gcd(int[] x) {
      if (x.length < 2) {
          throw new ArgumentException("Do not use this method if there are less than two numbers.");
      }
      int tmp = Gcd(x[x.length - 1], x[x.length - 2]);
      for (int i = x.length - 3; i >= 0; i--) {
          if (x[i] < 0) {
              throw new ArgumentException("Cannot compute the least common multiple of several numbers where one, at least, is negative.");
          }
          tmp = Gcd(tmp, x[i]);
      }
      return tmp;
  }

  public static int Gcd(int x1, int x2) {
      if (x1 < 0 || x2 < 0) {
          throw new ArgumentException("Cannot compute the GCD if one integer is negative.");
      }
      int a, b, g, z;

      if (x1 > x2) {
          a = x1;
          b = x2;
      } else {
          a = x2;
          b = x1;
      }

      if (b == 0) return 0;

      g = b;
      while (g != 0) {
          z= a % g;
          a = g;
          g = z;
      }
      return a;
  }

}

http://www.java2s.com/Tutorial/Java/0120__Development/GreatestCommonDivisorGCDofpositiveintegernumbers.htm

Wikipedia :

gcd是一个关联函数: gcd(a,gcd(b,c))=gcd(gcd(a,b),c)。

只需取前两个元素的gcd,然后计算结果和第三个元素的gcd,然后计算结果和第四个元素的gcd。。。

将其重写为单个函数。。。

    static int GCD(params int[] numbers)
    {
        Func<int, int, int> gcd = null;
        gcd = (a, b) => (b == 0 ? a : gcd(b, a % b));
        return numbers.Aggregate(gcd);
    } 

gcd(a1,a2,...,an)=gcd(a1,gcd(a2,gcd(a3...(gcd(a(n-1),an))))) gcd 计算结果为1。

gcd公司 gcd公司 计算结果为1,可以停止。

int GCD(int a,int b){ 
    return (!b) ? (a) : GCD(b, a%b);
}

void calc(a){
    int gcd = a[0];
    for(int i = 1 ; i < n;i++){
        if(gcd == 1){
            break;
        }
        gcd = GCD(gcd,a[i]);
    }
}

/*

Copyright (c) 2011, Louis-Philippe Lessard
All rights reserved.

Redistribution and use in source and binary forms, with or without modification, are permitted provided that the following conditions are met:

Redistributions of source code must retain the above copyright notice, this list of conditions and the following disclaimer.
Redistributions in binary form must reproduce the above copyright notice, this list of conditions and the following disclaimer in the documentation and/or other materials provided with the distribution.
THIS SOFTWARE IS PROVIDED BY THE COPYRIGHT HOLDERS AND CONTRIBUTORS "AS IS" AND ANY EXPRESS OR IMPLIED WARRANTIES, INCLUDING, BUT NOT LIMITED TO, THE IMPLIED WARRANTIES OF MERCHANTABILITY AND FITNESS FOR A PARTICULAR PURPOSE ARE DISCLAIMED. IN NO EVENT SHALL THE COPYRIGHT HOLDER OR CONTRIBUTORS BE LIABLE FOR ANY DIRECT, INDIRECT, INCIDENTAL, SPECIAL, EXEMPLARY, OR CONSEQUENTIAL DAMAGES (INCLUDING, BUT NOT LIMITED TO, PROCUREMENT OF SUBSTITUTE GOODS OR SERVICES; LOSS OF USE, DATA, OR PROFITS; OR BUSINESS INTERRUPTION) HOWEVER CAUSED AND ON ANY THEORY OF LIABILITY, WHETHER IN CONTRACT, STRICT LIABILITY, OR TORT (INCLUDING NEGLIGENCE OR OTHERWISE) ARISING IN ANY WAY OUT OF THE USE OF THIS SOFTWARE, EVEN IF ADVISED OF THE POSSIBILITY OF SUCH DAMAGE.

*/

unsigned gcd ( unsigned a, unsigned b );
unsigned gcd_arr(unsigned * n, unsigned size);

int main()
{
    unsigned test1[] = {8, 9, 12, 13, 39, 7, 16, 24, 26, 15};
    unsigned test2[] = {2, 4, 8, 16, 32, 64, 128, 256, 512, 1024, 2048};
    unsigned result;

    result = gcd_arr(test1, sizeof(test1) / sizeof(test1[0]));
    result = gcd_arr(test2, sizeof(test2) / sizeof(test2[0]));

    return result;
}


/**
* Find the greatest common divisor of 2 numbers
* See http://en.wikipedia.org/wiki/Greatest_common_divisor
*
* @param[in] a First number
* @param[in] b Second number
* @return greatest common divisor
*/
unsigned gcd ( unsigned a, unsigned b )
{
    unsigned c;
    while ( a != 0 )
    {
        c = a;
        a = b%a;
        b = c;
    }
    return b;
}

/**
* Find the greatest common divisor of an array of numbers
* See http://en.wikipedia.org/wiki/Greatest_common_divisor
*
* @param[in] n Pointer to an array of number
* @param[in] size Size of the array
* @return greatest common divisor
*/
unsigned gcd_arr(unsigned * n, unsigned size)
{
    unsigned last_gcd, i;
    if(size < 2) return 0;

    last_gcd = gcd(n[0], n[1]);

    for(i=2; i < size; i++)
    {
        last_gcd = gcd(last_gcd, n[i]);
    }

    return last_gcd;
}

Source code reference

这是最常用的三种方法:

public static uint FindGCDModulus(uint value1, uint value2)
{
    while(value1 != 0 && value2 != 0)
    {
            if (value1 > value2)
            {
                    value1 %= value2;
            }
            else
            {
                    value2 %= value1;
            }
    }
    return Math.Max(value1, value2);
       }

    public static uint FindGCDEuclid(uint value1, uint value2)
      {
    while(value1 != 0 && value2 != 0)
    {
            if (value1 > value2)
            {
                    value1 -= value2;
            }
            else
            {
                    value2 -= value1;
            }
    }
    return Math.Max(value1, value2);
  }

  public static uint FindGCDStein(uint value1, uint value2)
  {
    if (value1 == 0) return value2;
    if (value2 == 0) return value1;
    if (value1 == value2) return value1;

    bool value1IsEven = (value1 & 1u) == 0;
    bool value2IsEven = (value2 & 1u) == 0;

    if (value1IsEven && value2IsEven)
    {
            return FindGCDStein(value1 >> 1, value2 >> 1) << 1;
    }
    else if (value1IsEven && !value2IsEven)
    {
            return FindGCDStein(value1 >> 1, value2);
    }
    else if (value2IsEven)
    {
            return FindGCDStein(value1, value2 >> 1);
    }
    else if (value1 > value2)
    {
            return FindGCDStein((value1 - value2) >> 1, value2);
    }
    else
    {
            return FindGCDStein(value1, (value2 - value1) >> 1);
    }
  }

    static int GCD(int a, int b)
    {
        if (b == 0) return a;
        return GCD(b, a % b);
    }

    static int GCD(params int[] numbers)
    {
        int gcd = 0;
        int a = numbers[0];
        for(int i = 1; i < numbers.Length; i++)
        {
            gcd = GCD(a, numbers[i]);
            a = numbers[i];
        }

        return gcd;
    }

GCD(a,b,c)=GCD(a,GCD(b,c))=GCD(GCD(a,b),c)=GCD(GCD(a,c),b)

enter code here

公共类程序{static void Main(){控制台写入线(GCD(new[]{10,15,30,45}));}静态int GCD(int a,int b){返回b==0?a:GCD(b,a%b);}静态int GCD(int[]integerSet){返回集合整数集(GCD);}}

let a = 3
let b = 9

func gcd(a:Int, b:Int) -> Int {
    if a == b {
        return a
    }
    else {
        if a > b {
            return gcd(a:a-b,b:b)
        }
        else {
            return gcd(a:a,b:b-a)
        }
    }
}
print(gcd(a:a, b:b))

// pass all the values in array and call findGCD function
    int findGCD(int arr[], int n) 
    { 
        int gcd = arr[0]; 
        for (int i = 1; i < n; i++) {
            gcd = getGcd(arr[i], gcd); 
}

        return gcd; 
    } 

// check for gcd
int getGcd(int x, int y) 
    { 
        if (x == 0) 
            return y; 
        return gcd(y % x, x); 
    }