一组超过2个整数的最大公约数
Stack Overflow上有几个问题讨论如何找到两个值的最大公约数。 一个好的答案显示了一个简洁的递归函数来做到这一点。
但是如何找到一组超过2个整数的GCD？ 我似乎无法找到这样的例子。
任何人都可以建议最有效的代码来实现这个function吗？
static int GCD(int[] IntegerSet) { // what goes here? } 在这里，您可以使用LINQ和GCD方法编写代码示例。 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); }
你可以使用GCD的这个共同属性：
GCD(a, b, c) = GCD(a, GCD(b, c)) = GCD(GCD(a, b), c) = GCD(GCD(a, c), b)
假设你已经定义了GCD(a, b) ，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; } }
来源http://www.java2s.com/Tutorial/Java/0120__Development/GreatestCommonDivisorGCDofpositiveintegernumbers.htm
维基百科 ：
gcd是一个关联函数：gcd（a，gcd（b，c））= gcd（gcd（a，b），c）。
三个数字的gcd可以计算为gcd（a，b，c）= gcd（gcd（a，b），c），或者通过应用交换性和关联性以某种不同的方式计算。 这可以扩展到任意数量的数字。
只需取前两个元素的gcd，然后计算结果的gcd和第三个元素，然后计算结果的gcd和第四个元素……
将此重写为单个函数…
static int GCD(params int[] numbers) { Func 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))))) ，所以我会这样做如果一些gcd评估为1，则逐步中止。
如果您的数组已经排序，那么之前评估gcd对于较小的数字可能会更快，从那以后，一个gcd可能更有可能评估为1并且您可以停止。
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; }
源代码参考
这是最常用的三个：
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); } }
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; }
GCD（a，b，c）= GCD（a，GCD（b，c））= GCD（GCD（a，b），c）= GCD（GCD（a，c），b）
enter code here
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）; }}
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:ab,b:b) } else { return gcd(a:a,b:ba) } } } print(gcd(a:a, b:b))