blocks|key|32760|text|您的代码不计算输入数组的最大公约数-它计算有多少条目可以被数组中最小的元素d整除，然后有多少条目可以被一个较小的元素整除，依此类推，直到d为0。这与GCD一点关系都没有。|type|unstyled|depth|inlineStyleRanges|entityRanges|data|32761|一种简单的方法-尽管不一定是最快的-将基于以下事实:三个数字的GCD必须与其中任何一个数字的GCD相同，而其他两个数字的GCD必须相同。|32762|gcd(a,+b,+c)+=+gcd(gcd(a,+b),+c)+=+gcd(a,+gcd(b,+c))+=+gcd(gcd(a,+c),+b)|code-block|syntax|javascript|32763|扩展到n个输入是基本的：|32764|int+result+=+a[0];

for+(int+i+=+1;+i+<+a.Length;+%2B%2Bi)
++++result+=+gcd(result,+a[i]);|32765|在网上到处都可以找到两个数字的GCD的代码，例如在Rosetta+Code。我最喜欢的是这个简单的迭代版本：|offset|length|32766|int+gcd+(int+a,+int+b)
{
++++while+(b)
++++{
++++++++int+t+=+b;
++++++++b+=+a+%25+b;
++++++++a+=+t;
++++}

++++return+a;
}|32767|C#允许更简洁的公式，但在其他语言中可能不起作用(例如，在C%2B%2B中它会调用未定义的行为)：|32768|static+int+gcd+(int+a,+int+b)
{
++++while+(b+!=+0)
++++++++b+=+a+%25+(a+=+b);

++++return+a;
}|32769|entityMap|0|LINK|mutability|MUTABLE|url|https://rosettacode.org/wiki/Greatest_common_divisor#C^0|0|0|0|0|0|P|C|0|0|0|0|0^^$0|@$1|2|3|4|5|6|7|14|8|@]|9|@]|A|$]]|$1|B|3|C|5|6|7|15|8|@]|9|@]|A|$]]|$1|D|3|E|5|F|7|16|8|@]|9|@]|A|$G|H]]|$1|I|3|J|5|6|7|17|8|@]|9|@]|A|$]]|$1|K|3|L|5|F|7|18|8|@]|9|@]|A|$G|H]]|$1|M|3|N|5|6|7|19|8|@]|9|@$O|1A|P|1B|1|1C]]|A|$]]|$1|Q|3|R|5|F|7|1D|8|@]|9|@]|A|$G|H]]|$1|S|3|T|5|6|7|1E|8|@]|9|@]|A|$]]|$1|U|3|V|5|F|7|1F|8|@]|9|@]|A|$G|H]]|$1|W|3|-4|5|6|7|1G|8|@]|9|@]|A|$]]]|X|$Y|$5|Z|10|11|A|$12|13]]]]

Your code does not compute the greatest common divisor of the input array - it counts how many of the entries are evenly divisible by the smallest element d of the array, then how many are divisible by one smaller, and so on until d is 0. This has nothing to do with the GCD at all.

One easy way - though not necessarily the fastest - would be based on the fact that the GCD of three numbers must be the same as the GCD of any one of those numbers and the GCD of the other two. 

<pre><code>gcd(a, b, c) = gcd(gcd(a, b), c) = gcd(a, gcd(b, c)) = gcd(gcd(a, c), b)
</code></pre>

Extension to n inputs is elementary:

<pre><code>int result = a[0];

for (int i = 1; i &lt; a.Length; ++i)
 result = gcd(result, a[i]);
</code></pre>

Code for the GCD of two numbers can be found all over the 'net, for example at <a href="https://rosettacode.org/wiki/Greatest_common_divisor#C" rel="nofollow noreferrer">Rosetta Code</a>. One of my favourites is this plain iterative version:

<pre><code>int gcd (int a, int b)
{
 while (b)
 {
 int t = b;
 b = a % b;
 a = t;
 }

 return a;
}
</code></pre>

C# allows a more succinct formulation but in other languages this probably won't work (in C++ it would invoke undefined behaviour, for example):

<pre><code>static int gcd (int a, int b)
{
 while (b != 0)
 b = a % (a = b);

 return a;
}
</code></pre>

blocks|key|2494465|text|如果有些人觉得它有帮助，这里是欧几里德算法在JavaScript中的一个实现。|type|unstyled|depth|inlineStyleRanges|entityRanges|data|2494466|​|2494467|function+EuclideanGCD(a,+b)+{

++//+Make+sure+a+>+b,+interchange+values
++if+(a+<+b)+{
++++c+=+a;
++++a+=+b;
++++b+=+c
++}

++//+If+A+=+0+then+GCD(A,B)+=+B++and+we+can+stop.++
++if+(a+==+0)+{
++++return+b;

++//+If+B+=+0+then+GCD(A,B)+=+A++and+we+can+stop.++
++}+else+if+(b+==+0)+{
++++return+a;
++}+else+{

++++let+gdc+=+0;
++++let+quotient+=+Math.floor(a+/+b);+//+Get+the+divisor
++++let+remainder+=+a+%25+b;+//+Get+the+remainder

++++//+Make+a+recursive+call,+till+we+hit+0
++++gdc+=+EuclideanGCD(b,+remainder);
++++return+gdc;
++}

}

var+gcd+=+EuclideanGCD(234,+357);+
console.log(gcd);+//+Outputs:+3|code-block|syntax|javascript|2494468|2494469|entityMap^0|0|0|0|0^^$0|@$1|2|3|4|5|6|7|L|8|@]|9|@]|A|$]]|$1|B|3|C|5|6|7|M|8|@]|9|@]|A|$]]|$1|D|3|E|5|F|7|N|8|@]|9|@]|A|$G|H]]|$1|I|3|C|5|6|7|O|8|@]|9|@]|A|$]]|$1|J|3|-4|5|6|7|P|8|@]|9|@]|A|$]]]|K|$]]

In case some find it helpful, here is an implementation of the Euclidean algorithm in JavaScript.

<div class="snippet" data-lang="js" data-hide="false" data-console="true" data-babel="false">
<div class="snippet-code">
<pre class="snippet-code-js lang-js prettyprint-override"><code>function EuclideanGCD(a, b) {

 // Make sure a &gt; b, interchange values
 if (a &lt; b) {
 c = a;
 a = b;
 b = c
 }

 // If A = 0 then GCD(A,B) = B and we can stop. 
 if (a == 0) {
 return b;

 // If B = 0 then GCD(A,B) = A and we can stop. 
 } else if (b == 0) {
 return a;
 } else {

 let gdc = 0;
 let quotient = Math.floor(a / b); // Get the divisor
 let remainder = a % b; // Get the remainder

 // Make a recursive call, till we hit 0
 gdc = EuclideanGCD(b, remainder);
 return gdc;
 }

}

var gcd = EuclideanGCD(234, 357); 
console.log(gcd); // Outputs: 3</code></pre>
</div>
</div>

I know how to write a code finding a GCD of 2 number . However, I am trying to solve a problem of finding a GCD of n number and I think the algorithm is a little bit different than using an Eucledian algorithm. My code can be compiled , but it always gave me the wrong result. For example when i put n = 2 , GCD of 16 and 12 it gave the answer 8. Here is my code :

<pre><code>#include&lt;iostream&gt;
using namespace std;
int main()
{
 int a,b[100],c,d,e=0;
 cin&gt;&gt;a;
 for(c=0 ; c&lt;a ; c++)
 {
 cin&gt;&gt;b[c];
 }
 for(c=0 ; c&lt;a-1 ; c++)
 {
 if(c &lt; 1)
 {
 d = b[c]; 
 }
 if(b[c] &lt; d)
 {
 d = b[c];
 }
 }
 while(d&gt;0)
 {
 for(c=0 ; c&lt;a ; c++)
 {
 if(b[c] % d &lt; 1)
 {
 e++;
 }
 }
 if(e == c)
 {
 cout&lt;&lt;d;
 break;
 }
 d--;
 }
}
</code></pre>

Can you guys please find the mistake in my code?

GCD of Multiple Number

翻译质量差，导致语言生硬或混乱。

没有提供实际的解决方法或示例。

解答不清晰，无法理解或解决问题。

页面排版不美观，阅读体验差。

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我知道如何编写代码来查找2的GCD数字。然而，我正在尝试解决一个寻找n数的GCD的问题，我认为该算法与使用Eucledian算法略有不同。我的代码可以编译，但它总是给我错误的结果。例如，当我输入n = 2，GCD of 16和12时，结果是8。下面是我的代码：#include<iostream>using namesp...

问多个数字的GCD
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