使用Euclid算法查找GCF（GCD） [英] Using Euclid Algorithm to find GCF(GCD)

问题描述

我正在尝试使用Euclid算法（我发现在此。

I am trying to write a function to find the gcd of 2 numbers, using Euclid's Algorithm which I found here.

从较大的数字中减去尽可能小的数字，直到得到一个小于小数的数字。 （或没有得到否定的答案）现在，使用原始的小数字和结果（小数字）重复该过程。重复此过程，直到最后一个结果为零，并且GCF是倒数第二个小数结果。另请参见我们的Euclid算法计算器。

示例：找到GCF（18，27）

27-18 = 9

18-9 = 9

9-9 = 0

因此，18和27的最大公因数是9，在达到0之前我们得到的最小结果。

From the larger number, subtract the smaller number as many times as you can until you have a number that is smaller than the small number. (or without getting a negative answer) Now, using the original small number and the result, a smaller number, repeat the process. Repeat this until the last result is zero, and the GCF is the next-to-last small number result. Also see our Euclid's Algorithm Calculator.
Example: Find the GCF (18, 27)
27 - 18 = 9
18 - 9 = 9
9 - 9 = 0
So, the greatest common factor of 18 and 27 is 9, the smallest result we had before we reached 0.

按照以下说明，我编写了一个函数：

Following these instructions I wrote a function:

int hcf(int a, int b)
{
    int small = (a < b)? a : b;
    int big = (a > b)? a : b;
    int res;
    int gcf;

    cout << "small = " << small << "\n";
    cout << "big = " << big << "\n";

    while ((res = big - small) > small && res > 0) {
            cout << "res = " << res << "\n"; 
    }
    while ((gcf = small - res) > 0) {
        cout << "gcf = " << gcf << "\n";
    }


    return gcf;
}

但是，第二个循环似乎是无限的。谁能解释为什么？

However, the second loop seems to be infinite. Can anyone explain why?

我知道网站实际上显示了代码（PHP），但是我试图仅使用他们给出的说明来编写此代码。

I know the website actually shows the code(PHP), but I'm trying to write this code using only the instructions they give.

推荐答案

当然，此循环是无限的：

Of course this loop is infinite:

while ((gcf = small - res) > 0) {
    cout << "gcf = " << gcf << "\n";
}

small 和 res 不会在循环中更改，因此 gcf 也不会更改。该循环等效于：

small and res don't change in the loop, so gcf doesn't either. That loop is equivalent to:

gcf = small - res;
while (gcf > 0) {
    cout << "gcf = " << gcf << "\n";
}

这可能更清楚。

我会将该算法移植为以下代码：

I would port that algorithm to code as follows:

int gcd(int a, int b) {
    while (a != b) {
        if (a > b) {
            a -= b;
        }
        else {
            b -= a;
        }
    }
    return a;
}

虽然通常是 gcd 是使用mod来实现的，因为它要快得多：

Although typically gcd is implemented using mod, since it's much faster:

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

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