Java 解释互质检查是如何工作的

每次我用欧几里德方法检查两个数字是否是同素数。但是有一个解决方案使用了这个代码来检查共素数，这个方法所花费的时间比欧几里德方法要快得多：（）

我不明白这是怎么回事。（我确实理解位运算。）例如，这些行是什么意思

if (((u | v) & 1) == 0)
    return false;

---

为什么只返回false？还有其他一些我无法理解正在发生的事情。如果你们中有人能给我一些演练，这将非常有帮助。

您发布的代码是对的修改。以下是我的注释：

private static boolean isCoprime(long u, long v) {
    // If both numbers are even, then they are not coprime.
    if (((u | v) & 1) == 0) return false;

    // Now at least one number is odd. Eliminate all the factors of 2 from u.
    while ((u & 1) == 0) u >>= 1;

    // One is coprime with everything else by definition.
    if (u == 1) return true;

    do {
        // Eliminate all the factors of 2 from v, because we know that u and v do not have any 2's in common.
        while ((v & 1) == 0) v >>= 1;

        // One is coprime with everything else by definition.
        if (v == 1) return true;

        // Swap if necessary to ensure that v >= u.
        if (u > v) {
            long t = v;
            v = u;
            u = t;
        }

        // We know that GCD(u, v) = GCD(u, v - u).
        v -= u;
    } while (v != 0);

    // When we reach here, we have v = 0 and GCD(u, v) = current value of u, which is greater than 1.
    return false;
}

---

你能给我解释一下这个代码的另一部分吗。。如果可能的话？@user4890159还有什么其他部分？@user4890159在每次迭代中，你将

u

和

v

分开，直到两者都是奇数（或者

0

，也就是位移位魔法所做的），然后更新

u

和

v

。你可能想看看

private static boolean isCoprime(long u, long v) {
    // If both numbers are even, then they are not coprime.
    if (((u | v) & 1) == 0) return false;

    // Now at least one number is odd. Eliminate all the factors of 2 from u.
    while ((u & 1) == 0) u >>= 1;

    // One is coprime with everything else by definition.
    if (u == 1) return true;

    do {
        // Eliminate all the factors of 2 from v, because we know that u and v do not have any 2's in common.
        while ((v & 1) == 0) v >>= 1;

        // One is coprime with everything else by definition.
        if (v == 1) return true;

        // Swap if necessary to ensure that v >= u.
        if (u > v) {
            long t = v;
            v = u;
            u = t;
        }

        // We know that GCD(u, v) = GCD(u, v - u).
        v -= u;
    } while (v != 0);

    // When we reach here, we have v = 0 and GCD(u, v) = current value of u, which is greater than 1.
    return false;
}