我刚刚发现这个算法计算最大公约数在我的课堂笔记：

I just found this algorithm to compute the greatest common divisor in my lecture notes:

public static int gcd( int a, int b ) {
    while (b != 0) {
        final int r = a % b;
        a = b;
        b = r;
    }
    return a;
}

所以研究的是余数时分割的 B 的成的在的（获得MOD）。然后 B 的被分配给的一的，而其余被分配到的 B 的，并一的，则返回。我不能为我看到的生活是如何工作的！

So r is the remainder when dividing b into a (get the mod). Then b is assigned to a, and the remainder is assigned to b, and a is returned. I can't for the life of my see how this works!

和然后，显然该算法并不适用于所有的情况下工作，而这一次然后必须使用：

And then, apparently this algorithm doesn't work for all cases, and this one must then be used:

public static int gcd( int a, int b ) {
    final int gcd;
    if (b != 0) {
        final int q = a / b;
        final int r = a % b; // a == r + q * b AND r == a - q * b.
        gcd = gcd( b, r );
    } else {
        gcd = a;
    }
    return gcd;
}

我不明白这背后的原因。我一般得到递归和我精通Java，但是这是在暗指我。帮助吗？

I don't understand the reasoning behind this. I generally get recursion and am good at Java but this is alluding me. Help please?

推荐答案

维基百科的文章包含了一个解释，但它不容易立刻找到它（也，程序+防爆并不总是回答这个问题：为什么它的工作原理 ）。

The Wikipedia article contains an explanation, but it's not easy to find it immediately (also, procedure + proof don't always answer the question "why it works").

基本上它归结为这样一个事实：两个整数的a，b（假设> = b）中，它总是能写信= QB + r其中为r湾

Basically it comes down to the fact that for two integers a, b (assuming a >= b), it is always possible to write a = qb + r where r < b.

如果D = GCD（A，B），那么我们可以写一个= DS和b = DT。因此，我们有DS = QDT + R。因为左手侧是整除d时，右手侧也必须是整除用d。而且，由于QDT是整除研发，得出的结论是，其中r还必须除尽用d。

If d=gcd(a,b) then we can write a=ds and b=dt. So we have ds = qdt + r. Since the left hand side is divisible by d, the right hand side must also be divisible by d. And since qdt is divisible by d, the conclusion is that r must also be divisible by d.

要总结一下：我们有一个= BQ + R，其中R; b和a，b和r均为所有整除最大公约数（A，B）。

To summarise: we have a = bq + r where r < b and a, b and r are all divisible by gcd(a,b).

由于A> = B> R，我们有两种情况：

Since a >= b > r, we have two cases:

为什么这是一个减少？由于R＆LT;湾因此，我们现在处理的绝对数量较少。这意味着，我们只有在我们到达R = 0申请本次减持的次数是有限的。

Why is this a reduction? Because r < b. So we are dealing with numbers that are definitely smaller. This means that we only have to apply this reduction a finite number of times before we reach r = 0.

现在，R = A％B的希望解释了code你。

Now, r = a % b which hopefully explains the code you have.