我需要做以下算术：
long a,b,c;
long result = a*b/c;
虽然结果保证适合
long我试图一步一步地做到这一点（先乘法然后除法），同时通过将
a*bint[] mag在这里我陷入了分裂。我无法理解进行精确除法所需的按位移位。我只需要商（不需要余数）。
假设的方法是：
public static long divide(int[] dividend, long divisor)
另外，我不考虑使用
BigInteger任何帮助将不胜感激！
编辑： 我并不是想自己实现整个
BigIntegera*b/ca*bBigIntegerEdit2：如果能以一种聪明的方式完成，完全不溢出，那就太理想了，评论中出现了一些提示，但我仍在寻找正确的提示。
更新： 我尝试将 BigInteger 代码移植到我的特定需求，而不创建对象，并且在第一次迭代中，与使用 BigInteger（在我的开发电脑上）相比，我的速度提高了约 46%。
然后我尝试了一些修改过的@David Eisenstat解决方案，与BigInteger（即〜18％）相比，这给了我〜56％（我从
Long.MIN_VALUELong.MAX_VALUE还会有更多的优化和测试迭代，但在这一点上，我想我必须接受这个答案是最好的。
我一直在修改一种方法，即 (1) 将
abca*b - c*qdoublelongpublic 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 != %s\n", 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;
}
您可以使用最大公约数（gcd）来帮助。
a * b / c = (a / gcd(a,c)) * (b / (c / gcd(a,c)))
编辑：OP要求我解释上面的方程。基本上，我们有：
a = (a / gcd(a,c)) * gcd(a,c)
c = (c / gcd(a,c)) * gcd(a,c)
Let's say x=gcd(a,c) for brevity, and rewrite this.
a*b/c = (a/x) * x * b
--------------
(c/x) * x
Next, we cancel
a*b/c = (a/x) * b
----------
(c/x)
您可以更进一步。设 y = gcd(b, c/x)
a*b/c = (a/x) * (b/y) * y
------------------
((c/x)/y) * y
a*b/c = (a/x) * (b/y)
------------
(c/(xy))
这是获取 gcd 的代码。
static long gcd(long a, long b)
{
if (b == 0)
return a;
return gcd(b, a % b);
}
David Eisenstat 让我思考更多。
我希望简单的情况能够快速完成：让
double /** Multiplies both <code>factor</code>s
* and divides by <code>divisor</code>.
* @return <code>Long.MIN_VALUE</code> if result out of range,<br/>
* else <code>factorA * factor1 / divisor</code> */
public static long
mulDiv(long factorA, long factor1, long divisor) {
final double dd = divisor,
product = (double)factorA * factor1,
a1_d = product / dd;
if (a1_d < -TOO_LARGE || TOO_LARGE < a1_d)
return tooLarge();
if (-ONE_ < a1_d && a1_d < ONE_)
return 0;
if (-EXACT < product && product < EXACT)
return (long) a1_d;
long pLo = factorA * factor1, //diff,
pHi = high64(factorA, factor1);
if (a1_d < -LONG_MAX_ || LONG_MAX_ < a1_d) {
long maxdHi = divisor >> 1;
if (maxdHi < pHi
|| maxdHi == pHi
&& Long.compareUnsigned((divisor << Long.SIZE-1),
pLo) <= 0)
return tooLarge();
}
final double high_dd = TWO_POWER64/dd;
long quotient = (long) a1_d,
loPP = quotient * divisor,
hiPP = high64(quotient, divisor);
long remHi = pHi - hiPP, // xxx overflow/carry
remLo = pLo - loPP;
if (Long.compareUnsigned(pLo, remLo) < 0)
remHi -= 1;
double fudge = remHi * high_dd;
if (remLo < 0)
fudge += high_dd;
fudge += remLo/dd;
long //fHi = (long)fudge/TWO_POWER64,
fLo = (long) Math.floor(fudge); //*round
quotient += fLo;
loPP = quotient * divisor;
hiPP = high64(quotient, divisor);
remHi = pHi - hiPP; // should be 0?!
remLo = pLo - loPP;
if (Long.compareUnsigned(pLo, remLo) < 0)
remHi -= 1;
if (0 == remHi && 0 <= remLo && remLo < divisor)
return quotient;
fudge = remHi * high_dd;
if (remLo < 0)
fudge += high_dd;
fudge += remLo/dd;
fLo = (long) Math.floor(fudge);
return quotient + fLo;
}
/** max <code>double</code> trusted to represent
* a value in the range of <code>long</code> */
static final double
LONG_MAX_ = Double.valueOf(Long.MAX_VALUE - 0xFFF);
/** max <code>double</code> trusted to represent a value below 1 */
static final double
ONE_ = Double.longBitsToDouble(
Double.doubleToRawLongBits(1) - 4);
/** max <code>double</code> trusted to represent a value exactly */
static final double
EXACT = Long.MAX_VALUE >> 12;
static final double
TWO_POWER64 = Double.valueOf(1L<<32)*Double.valueOf(1L<<32);
static long tooLarge() {
// throw new RuntimeException("result too large for long");
return Long.MIN_VALUE;
}
static final long ONES_32 = ~(~0L << 32);
static long high64(long factorA, long factor1) {
long loA = factorA & ONES_32,
hiA = factorA >>> 32,
lo1 = factor1 & ONES_32,
hi1 = factor1 >>> 32;
return ((loA * lo1 >>> 32)
+loA * hi1 + hiA * lo1 >>> 32)
+ hiA * hi1;
}
（我在 IDE 之外重新排列了这段代码，使
mulDiv()也许不聪明，但结果时间是线性的
#define MUL_DIV_TYPE unsigned int
#define BITS_PER_TYPE (sizeof(MUL_DIV_TYPE)*8)
#define TOP_BIT_TYPE (1<<(BITS_PER_TYPE-1))
//
// result = ( a * b ) / c, without intermediate overflow.
//
MUL_DIV_TYPE mul_div( MUL_DIV_TYPE a, MUL_DIV_TYPE b, MUL_DIV_TYPE c ) {
MUL_DIV_TYPE st, sb; // product sum top and bottom
MUL_DIV_TYPE d, e; // division result
MUL_DIV_TYPE i, // bit counter
j; // overflow check
st = 0;
sb = 0;
d = 0;
e = 0;
for( i = 0; i < BITS_PER_TYPE; i++ ) {
//
// Shift sum left to make space
// for next partial sum
//
st <<= 1;
if( sb & TOP_BIT_TYPE ) st |= 1;
sb <<= 1;
//
// Add a to s if top bit on b
// is set.
//
if( b & TOP_BIT_TYPE ) {
j = sb;
sb += a;
if( sb < j ) st++;
}
//
// Division.
//
d <<= 1;
if( st >= c ) {
d |= 1;
st -= c;
e++;
}
else {
if( e ) e++;
}
//
// Shift b up by one bit.
//
b <<= 1;
}
//
// Roll in missing bits.
//
for( i = e; i < BITS_PER_TYPE; i++ ) {
//
// Shift across product sum
//
st <<= 1;
if( sb & TOP_BIT_TYPE ) st |= 1;
sb <<= 1;
//
// Division, continued.
//
d <<= 1;
if( st >= c ) {
d |= 1;
st -= c;
}
}
return( d ); // remainder should be in st
}
我创建了以下看起来相当不错的算法：
BigIntegerlong /**
* Long computation of (a*b)/c. Arguments and result must be in range 0..Long.MAX_VALUE
*
* @param a First multiplicand
* @param b Second multiplicand
* @param c Divisor
* @return Result of (a*b)/c, or -1 if result overflows long
*/
static long fastMulDiv(long a, long b, long c) {
// 128 bit multiply
long m0=a*b;
long m1=Math.multiplyHigh(a, b);
long acc=0;
// we are going to do base 2^63 long division :-)
// a * b = ab1 * 2^63 + ab0
long ab1 = (m1<<1)|(m0>>>63);
if (c<=ab1) return -1; // we know this will overflow
if (ab1==0) return m0/c; // fast path
long ab0 = (m0&~0x8000000000000000L);
// d = 2^63 / c
long dq=-(0x8000000000000000L/c); // note we need to reverse sign
long dr=Long.remainderUnsigned(0x8000000000000000L, c); // dr < c
while (ab1>0) {
// a * b = c*(ab1*dq) + (ab1*dr + ab0)
acc+=ab1*dq;
// so we just need to divide (ab1*dr + ab0) by c to get the rest
m0=ab1*dr;
m1=Math.multiplyHigh(ab1, dr);
ab1 = (m1<<1)|(m0>>>63);
ab0 = (m0&~0x8000000000000000L)+ab0; // note we add in the previous ab0
if (ab0<0) {
// overflow carry
ab0=(ab0&~0x8000000000000000L);
ab1++;
}
}
long result = acc + ab0/c;
return result;
}
将 a/c 和 b/c 分成整数部分和分数（余数）部分，然后你有：
a*b/c
= c * a/c * b/c
= c * (x/c + y/c) * (z/c + w/c)
= xz/c + xw/c + yz/c + yw/c where x and z are multiples of c
因此，您可以轻松计算前三个因子而不会溢出。根据我的经验，这通常足以涵盖典型的溢出情况。然而，如果你的 divisor 太大，导致
(a % c) * (b % c)无论如何，上面的代码：
long a,b,c;
long bMod = (b % c)
long result = a * (b / c) + (a / c) * bMod + ((a % c) * bMod) / c;
如果速度是一个大问题（我假设它至少在某种程度上是一个大问题，因为你问这个），你可能需要考虑将
a/cb/c(a % c)(a - aDiv * c)