我一直在尝试一种方法（1）进行乘法运算，
a并
b使用21位四肢上的school算法（2）进行长除法运算，并使用a
来存储高阶位
c的残差的不寻常表示形式并存储低位。我不知道是否可以与标准的长距离比赛竞争，但是为了您的享受，
a*b -c*q``double``long
public class MulDiv { public static void main(String[] args) { java.util.Random r = new java.util.Random(); for (long i = 0; true; i++) { if (i % 1000000 == 0) { System.err.println(i); } long a = r.nextLong() >> (r.nextInt(8) * 8); long b = r.nextLong() >> (r.nextInt(8) * 8); long c = r.nextLong() >> (r.nextInt(8) * 8); if (c == 0) { continue; } long x = mulDiv(a, b, c); java.math.BigInteger aa = java.math.BigInteger.valueOf(a); java.math.BigInteger bb = java.math.BigInteger.valueOf(b); java.math.BigInteger cc = java.math.BigInteger.valueOf(c); java.math.BigInteger xx = aa.multiply(bb).divide(cc); if (java.math.BigInteger.valueOf(xx.longValue()).equals(xx) && x != xx.longValue()) { System.out.printf("a=%d b=%d c=%d: %d != %sn", a, b, c, x, xx); } } } // Returns truncate(a b/c), subject to the precondition that the result is // defined and can be represented as a long. private static long mulDiv(long a, long b, long c) { // Decompose a. long a2 = a >> 42; long a10 = a - (a2 << 42); long a1 = a10 >> 21; long a0 = a10 - (a1 << 21); assert a == (((a2 << 21) + a1) << 21) + a0; // Decompose b. long b2 = b >> 42; long b10 = b - (b2 << 42); long b1 = b10 >> 21; long b0 = b10 - (b1 << 21); assert b == (((b2 << 21) + b1) << 21) + b0; // Compute a b. long ab4 = a2 * b2; long ab3 = a2 * b1 + a1 * b2; long ab2 = a2 * b0 + a1 * b1 + a0 * b2; long ab1 = a1 * b0 + a0 * b1; long ab0 = a0 * b0; // Compute a b/c. DivBy d = new DivBy(c); d.shift21Add(ab4); d.shift21Add(ab3); d.shift21Add(ab2); d.shift21Add(ab1); d.shift21Add(ab0); return d.getQuotient(); }}public strictfp class DivBy { // Initializes n <- 0. public DivBy(long d) { di = d; df = (double) d; oneOverD = 1.0 / df; } // Updates n <- 2^21 n + i. Assumes |i| <= 3 (2^42). public void shift21Add(long i) { // Update the quotient and remainder. q <<= 21; ri = (ri << 21) + i; rf = rf * (double) (1 << 21) + (double) i; reduce(); } // Returns truncate(n/d). public long getQuotient() { while (rf != (double) ri) { reduce(); } // Round toward zero. if (q > 0) { if ((di > 0 && ri < 0) || (di < 0 && ri > 0)) { return q - 1; } } else if (q < 0) { if ((di > 0 && ri > 0) || (di < 0 && ri < 0)) { return q + 1; } } return q; } private void reduce() { // x is approximately r/d. long x = Math.round(rf * oneOverD); q += x; ri -= di * x; rf = repairLowOrderBits(rf - df * (double) x, ri); } private static double repairLowOrderBits(double f, long i) { int e = Math.getExponent(f); if (e < 64) { return (double) i; } long rawBits = Double.doubleToRawLongBits(f); long lowOrderBits = (rawBits >> 63) ^ (rawBits << (e - 52)); return f + (double) (i - lowOrderBits); } private final long di; private final double df; private final double oneOverD; private long q = 0; private long ri = 0; private double rf = 0;}
