从一组2个以上整数最大公约数

有堆栈溢出几个问题讨论如何找到两个值的最大公约数。 一个很好的回答显示了一个整洁的递归函数来做到这一点。

但我怎么能找到一套超过2个整数的GCD? 我似乎无法找到这样的一个例子。



任何人都可以建议实现该功能的最有效的代码?

static int GCD(int[] IntegerSet) { // what goes here? }

--------------解决方案-------------

在这里,你必须利用问题,你链接的LINQ和GCD方法的代码示例。 它是利用在其他的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);
}

你可以使用一个GCD这个共同属性:

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

假设你有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);
}
}

下面是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;
}

}

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

维基百科:

最大公约数是一个关联函数:gcd上述(一,GCD(B,C))= GCD(GCD(A,B),C)。

三个数字的最大公约数可以计算为GCD(A,B,C)=最大公约数(GCD(A,B),c)中,或者在通过施加交换性和结合一些不同的方式。 这可以扩展到任何数量的数字。

只取前两个元素的GCD,然后计算出结果的GCD和第三个元素,然后计算结果和第四个元素的GCD ...

重写此作为一个单一的功能...

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)))))所以我这样做只是一步中止,如果一些步骤gcd的计算结果为1。

如果你的数组排序,它可能会更快评估gcd的少数较早,自那时以来,它可能更容易,一个gcd计算结果为1,你可以停止。

/*

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;
}

源代码参考

这是三个最常见的使用:

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);
}
}

如果不使用LINQ。

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;
}

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]);
}
}

分类:C# 时间:2013-03-28 人气:0
本文关键词: 算法,C#,数学
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