以下函数可用于(相对快速地)将64位无符号整数分解为其素因子。注意,因子分解不是概率性的(即,它是精确的)。在现代硬件上,该算法已经足够快,可以在几秒钟内找到一个数字是素数或几乎没有很大的因子。
问题: 可以对所提出的算法进行任何改进,同时保持其单线程,以便它可以更快地考虑(任意)非常大的无符号64位整数,优选地不使用概率方法(例如,Miller-Rabin)来确定素数?
// system specific typedef for ulong should go here (or use boost::uint64_t)
typedef unsigned __int64 ulong;
typedef std::vector<ulong> ULongVector;
// Caller needs to pass in an empty factors vector
void GetFactors(ULongVector &factors, ulong num)
{
// Num has to be at least 2 to contain "prime" factors
if (num<2)
return;
ulong workingNum=num;
ulong nextOffset=2; // Will be used to skip multiples of 3, later
// Factor out factors of 2
while (workingNum%2==0)
{
factors.push_back(2);
workingNum/=2;
}
// Factor out factors of 3
while (workingNum%3==0)
{
factors.push_back(3);
workingNum/=3;
}
// If all of the factors were 2s and 3s, done...
if (workingNum==1)
return;
// sqrtNum is the (inclusive) upper bound of our search for factors
ulong sqrtNum=(ulong) sqrt(double(workingNum+0.5));
// Factor out potential factors that are greate than or equal to 5
// The variable n represents the next potential factor to be tested
for (ulong n=5;n<=sqrtNum;)
{
// Is n a factor of the current working number?
if (workingNum%n==0)
{
// n is a factor, so add it to the list of factors
factors.push_back(n);
// Divide current working number by n, to get remaining number to factor
workingNum/=n;
// Check if we've found all factors
if (workingNum==1)
return;
// Recalculate the new upper bound for remaining factors
sqrtNum=(ulong) sqrt(double(workingNum+0.5));
// Recheck if n is a factor of the new working number,
// in case workingNum contains multiple factors of n
continue;
}
// n is not or is no longer a factor, try the next odd number
// that is not a multiple of 3
n+=nextOffset;
// Adjust nextOffset to be an offset from n to the next non-multiple of 3
nextOffset=(nextOffset==2UL ? 4UL : 2UL);
}
// Current workingNum is prime, add it as a factor
factors.push_back(workingNum);
}
由于
编辑:我添加了更多评论。向量通过引用传入的原因是允许向量在调用之间重用,并避免动态分配。向量未在函数中清空的原因是允许将当前“num”因子附加到向量中已有的因子的奇怪要求。
函数本身并不漂亮,可以重构,但问题是如何使算法更快。所以,请不要提出如何使函数更漂亮,可读或C ++的建议。这是孩子的游戏。改进这种算法,以便能够更快地找到(证明的)因素是困难的部分。
更新: Potatoswatter 到目前为止还有一些出色的解决方案,请务必查看底部附近的MMX解决方案。
答案 0 :(得分:19)
将这种方法与(预先生成的)筛子进行比较。模数很昂贵,因此这两种方法基本上都做了两件事:生成潜在因素,并执行模运算。任何一个程序都应该以比模数更少的周期合理地生成一个新的候选因子,因此任何一个程序都是模数约束的。
给定的方法过滤掉所有整数的恒定比例,即2和3的倍数,或75%。四分之一(给定)数字用作模运算符的参数。我称之为跳过滤镜。
另一方面,筛子仅使用素数作为模运算符的参数,并且平均difference between successive primes由prime number theorem控制为1 / ln(N)。例如, e ^ 20只有不到5亿,所以超过5亿的人有5%的机会成为素数。如果考虑所有高达2 ^ 32的数字,5%是一个很好的经验法则。
因此,筛选将花费5倍于div
操作作为跳过筛选器的时间。要考虑的下一个因素是筛子产生质数的速度,即从存储器或磁盘读取它们的速度。如果获取一个素数比4 div
s快,则筛子更快。根据我的表div
,我的Core2上的吞吐量最多每12个周期一个。这些将是严格的划分问题,所以让我们保守地预算每个素数50个周期。对于2.5 GHz处理器,这是20纳秒。
在20 ns内,50 MB /秒的硬盘驱动器可以读取大约一个字节。简单的解决方案是每个素数使用4个字节,因此驱动器会更慢。但是,我们可以更聪明。如果我们想按顺序编码所有素数,我们可以只编码它们的差异。同样,预期的差异是1 / ln(N)。而且,它们都是均匀的,这节省了额外的一点。它们永远不会为零,这使得多字节编码的扩展免费。因此,每个素数使用一个字节,最多512个差异可以存储在一个字节中,根据that Wikipedia article,我们最多可以达到303371455241。
因此,根据硬盘驱动器,存储的素数列表在验证素数时的速度应该大致相等。如果它可以存储在RAM中(它是203 MB,因此后续运行可能会达到磁盘缓存),那么问题就完全消失了,因为FSB速度通常与处理器速度相差小于FSB宽度(以字节为单位) - 即,FSB每个周期可以传输多个素数。然后,改进因素是除法运算的减少,即五次。下面的实验结果证实了这一点。
当然,那就是多线程。可以将素数或跳过过滤候选者的范围分配给不同的线程,使得任一方法都令人尴尬地平行。没有优化不涉及增加并行分频器电路的数量,除非你以某种方式消除模数。
这是一个这样的程序。这是模板化的,所以你可以添加bignums。
/*
* multibyte_sieve.cpp
* Generate a table of primes, and use it to factorize numbers.
*
* Created by David Krauss on 10/12/10.
*
*/
#include <cmath>
#include <bitset>
#include <limits>
#include <memory>
#include <fstream>
#include <sstream>
#include <iostream>
#include <iterator>
#include <stdint.h>
using namespace std;
char const primes_filename[] = "primes";
enum { encoding_base = (1<< numeric_limits< unsigned char >::digits) - 2 };
template< typename It >
unsigned decode_gap( It &stream ) {
unsigned gap = static_cast< unsigned char >( * stream ++ );
if ( gap ) return 2 * gap; // only this path is tested
gap = ( decode_gap( stream )/2-1 ) * encoding_base; // deep recursion
return gap + decode_gap( stream ); // shallow recursion
}
template< typename It >
void encode_gap( It &stream, uint32_t gap ) {
unsigned len = 0, bytes[4];
gap /= 2;
do {
bytes[ len ++ ] = gap % encoding_base;
gap /= encoding_base;
} while ( gap );
while ( -- len ) { // loop not tested
* stream ++ = 0;
* stream ++ = bytes[ len + 1 ];
}
* stream ++ = bytes[ 0 ];
}
template< size_t lim >
void generate_primes() {
auto_ptr< bitset< lim / 2 > > sieve_p( new bitset< lim / 2 > );
bitset< lim / 2 > &sieve = * sieve_p;
ofstream out_f( primes_filename, ios::out | ios::binary );
ostreambuf_iterator< char > out( out_f );
size_t count = 0;
size_t last = sqrtl( lim ) / 2 + 1, prev = 0, x = 1;
for ( ; x != last; ++ x ) {
if ( sieve[ x ] ) continue;
size_t n = x * 2 + 1; // translate index to number
for ( size_t m = x + n; m < lim/2; m += n ) sieve[ m ] = true;
encode_gap( out, ( x - prev ) * 2 );
prev = x;
}
for ( ; x != lim / 2; ++ x ) {
if ( sieve[ x ] ) continue;
encode_gap( out, ( x - prev ) * 2 );
prev = x;
}
cout << prev * 2 + 1 << endl;
}
template< typename I >
void factorize( I n ) {
ifstream in_f( primes_filename, ios::in | ios::binary );
if ( ! in_f ) {
cerr << "Could not open primes file.\n"
"Please generate it with 'g' command.\n";
return;
}
while ( n % 2 == 0 ) {
n /= 2;
cout << "2 ";
}
unsigned long factor = 1;
for ( istreambuf_iterator< char > in( in_f ), in_end; in != in_end; ) {
factor += decode_gap( in );
while ( n % factor == 0 ) {
n /= factor;
cout << factor << " ";
}
if ( n == 1 ) goto finish;
}
cout << n;
finish:
cout << endl;
}
int main( int argc, char *argv[] ) {
if ( argc != 2 ) goto print_help;
unsigned long n;
if ( argv[1][0] == 'g' ) {
generate_primes< (1ul<< 32) >();
} else if ( ( istringstream( argv[1] ) >> n ).rdstate() == ios::eofbit )
factorize( n );
} else goto print_help;
return 0;
print_help:
cerr << "Usage:\n\t" << argv[0] << " <number> -- factorize number.\n"
"\t" << argv[0] << " g -- generate primes file in current directory.\n";
}
2.2 GHz MacBook Pro的性能:
dkrauss$ time ./multibyte_sieve g
4294967291
real 2m8.845s
user 1m15.177s
sys 0m2.446s
dkrauss$ time ./multibyte_sieve 18446743721522234449
4294967231 4294967279
real 0m5.405s
user 0m4.773s
sys 0m0.458s
dkrauss$ time ./mike 18446743721522234449
4294967231 4294967279
real 0m25.147s
user 0m24.170s
sys 0m0.096s
答案 1 :(得分:9)
我的另一个答案是相当长的,与此完全不同,所以这里还有别的东西。
该程序不是仅过滤掉前两个素数的倍数,或者每个字节编码所有相关的素数,而是过滤掉所有适合8位的素数的倍数,特别是2到211.所以不要通过33%的数字,这大约10%用于分部运营商。
质数保存在三个SSE寄存器中,其运行计数器的模数保持在另外三个。如果具有计数器的任何素数的模数等于零,则计数器不能是素数。此外,如果任何模数等于1,则(计数器+ 2)不能是素数等,直到(计数器+30)。偶数会被忽略,并且会跳过+3,+ 6和+5等偏移量。向量处理允许一次更新16个字节大小的变量。
在抛出一个完全微观优化的厨房水槽之后(但没有任何平台特定于内联指令),我的性能提升了1.78倍(我的笔记本电脑上的24秒对13.4秒)。如果使用bignum库(即使是非常快的库),优势也更大。请参阅下面的更具可读性的预优化版本。
/*
* factorize_sse.cpp
* Filter out multiples of the first 47 primes while factorizing a number.
*
* Created by David Krauss on 10/14/10.
*
*/
#include <cmath>
#include <sstream>
#include <iostream>
#include <xmmintrin.h>
using namespace std;
inline void remove_factor( unsigned long &n, unsigned long factor ) __attribute__((always_inline));
void remove_factor( unsigned long &n, unsigned long factor ) {
while ( n % factor == 0 ) {
n /= factor;
cout << factor << " ";
}
}
int main( int argc, char *argv[] ) {
unsigned long n;
if ( argc != 2
|| ( istringstream( argv[1] ) >> n >> ws ).rdstate() != ios::eofbit ) {
cerr << "Usage: " << argv[0] << " <number>\n";
return 1;
}
int primes[] = { 2, 3, 5, 7, 11, 13, 17, 19, 23, 29, 31, 37, 41, 43, 47,
53, 59, 61, 67, 71, 73, 79, 83, 89, 97, 101, 103, 107, 109, 113, 127,
131, 137, 139, 149, 151, 157, 163, 167, 173, 179, 181, 191, 193, 197, 199, 211 };
for ( int *p = primes; p < primes + sizeof primes/sizeof *primes; ++ p ) {
remove_factor( n, * p );
}
//int histo[8] = {}, total = 0;
enum { bias = 15 - 128 };
__m128i const prime1 = _mm_set_epi8( 21, 21, 21, 22, 22, 26, 26, 17, 19, 23, 29, 31, 37, 41, 43, 47 ),
prime2 = _mm_set_epi8( 53, 59, 61, 67, 71, 73, 79, 83, 89, 97, 101, 103, 107, 109, 113, 127 ),
prime3 = _mm_set_epi8( 131, 137, 139, 149, 151, 157, 163, 167, 173, 179, 181, 191, 193, 197, 199, 211 ),
vbias = _mm_set1_epi8( bias ),
v3 = _mm_set1_epi8( 3+bias ), v5 = _mm_set1_epi8( 5+bias ), v6 = _mm_set1_epi8( 6+bias ), v8 = _mm_set1_epi8( 8+bias ),
v9 = _mm_set1_epi8( 9+bias ), v11 = _mm_set1_epi8( 11+bias ), v14 = _mm_set1_epi8( 14+bias ), v15 = _mm_set1_epi8( 15+bias );
__m128i mod1 = _mm_add_epi8( _mm_set_epi8( 3, 10, 17, 5, 16, 6, 19, 8, 9, 11, 14, 15, 18, 20, 21, 23 ), vbias ),
mod2 = _mm_add_epi8( _mm_set_epi8( 26, 29, 30, 33, 35, 36, 39, 41, 44, 48, 50, 51, 53, 54, 56, 63 ), vbias ),
mod3 = _mm_add_epi8( _mm_set_epi8( 65, 68, 69, 74, 75, 78, 81, 83, 86, 89, 90, 95, 96, 98, 99, 105 ), vbias );
for ( unsigned long factor = 1, limit = sqrtl( n ); factor <= limit + 30; factor += 30 ) {
if ( n == 1 ) goto done;
// up to 2^32, distribution of number candidates produced (0 up to 7) is
// 0.010841 0.0785208 0.222928 0.31905 0.246109 0.101023 0.0200728 0.00145546
unsigned candidates[8], *cand_pen = candidates;
* cand_pen = 6;
cand_pen += !( _mm_movemask_epi8( _mm_cmpeq_epi8( mod1, v3 ) ) | _mm_movemask_epi8( _mm_or_si128( _mm_cmpeq_epi8( mod2, v3 ), _mm_cmpeq_epi8( mod3, v3 ) ) ) );
* cand_pen = 10;
cand_pen += !( _mm_movemask_epi8( _mm_cmpeq_epi8( mod1, v5 ) ) | _mm_movemask_epi8( _mm_or_si128( _mm_cmpeq_epi8( mod2, v5 ), _mm_cmpeq_epi8( mod3, v5 ) ) ) );
* cand_pen = 12;
cand_pen += !( _mm_movemask_epi8( _mm_cmpeq_epi8( mod1, v6 ) ) | _mm_movemask_epi8( _mm_or_si128( _mm_cmpeq_epi8( mod2, v6 ), _mm_cmpeq_epi8( mod3, v6 ) ) ) );
* cand_pen = 16;
cand_pen += !( _mm_movemask_epi8( _mm_cmpeq_epi8( mod1, v8 ) ) | _mm_movemask_epi8( _mm_or_si128( _mm_cmpeq_epi8( mod2, v8 ), _mm_cmpeq_epi8( mod3, v8 ) ) ) );
* cand_pen = 18;
cand_pen += !( _mm_movemask_epi8( _mm_cmpeq_epi8( mod1, v9 ) ) | _mm_movemask_epi8( _mm_or_si128( _mm_cmpeq_epi8( mod2, v9 ), _mm_cmpeq_epi8( mod3, v9 ) ) ) );
* cand_pen = 22;
cand_pen += !( _mm_movemask_epi8( _mm_cmpeq_epi8( mod1, v11 ) ) | _mm_movemask_epi8( _mm_or_si128( _mm_cmpeq_epi8( mod2, v11 ), _mm_cmpeq_epi8( mod3, v11 ) ) ) );
* cand_pen = 28;
cand_pen += !( _mm_movemask_epi8( _mm_cmpeq_epi8( mod1, v14 ) ) | _mm_movemask_epi8( _mm_or_si128( _mm_cmpeq_epi8( mod2, v14 ), _mm_cmpeq_epi8( mod3, v14 ) ) ) );
* cand_pen = 30;
cand_pen += !( _mm_movemask_epi8( _mm_cmpeq_epi8( mod1, v15 ) ) | _mm_movemask_epi8( _mm_or_si128( _mm_cmpeq_epi8( mod2, v15 ), _mm_cmpeq_epi8( mod3, v15 ) ) ) );
/*++ total;
++ histo[ cand_pen - candidates ];
cout << "( ";
while ( cand_pen != candidates ) cout << factor + * -- cand_pen << " ";
cout << ")" << endl; */
mod1 = _mm_sub_epi8( mod1, _mm_set1_epi8( 15 ) ); // update residuals
__m128i mask1 = _mm_cmplt_epi8( mod1, _mm_set1_epi8( 1+bias ) );
mask1 = _mm_and_si128( mask1, prime1 ); // residual goes to zero or negative?
mod1 = _mm_add_epi8( mask1, mod1 ); // combine reset into zero or negative
mod2 = _mm_sub_epi8( mod2, _mm_set1_epi8( 15 ) );
__m128i mask2 = _mm_cmplt_epi8( mod2, _mm_set1_epi8( 1+bias ) );
mask2 = _mm_and_si128( mask2, prime2 );
mod2 = _mm_add_epi8( mask2, mod2 );
mod3 = _mm_sub_epi8( mod3, _mm_set1_epi8( 15 ) );
__m128i mask3 = _mm_cmplt_epi8( mod3, _mm_set1_epi8( 1+bias ) );
mask3 = _mm_and_si128( mask3, prime3 );
mod3 = _mm_add_epi8( mask3, mod3 );
if ( cand_pen - candidates == 0 ) continue;
remove_factor( n, factor + candidates[ 0 ] );
if ( cand_pen - candidates == 1 ) continue;
remove_factor( n, factor + candidates[ 1 ] );
if ( cand_pen - candidates == 2 ) continue;
remove_factor( n, factor + candidates[ 2 ] );
if ( cand_pen - candidates == 3 ) continue;
remove_factor( n, factor + candidates[ 3 ] );
if ( cand_pen - candidates == 4 ) continue;
remove_factor( n, factor + candidates[ 4 ] );
if ( cand_pen - candidates == 5 ) continue;
remove_factor( n, factor + candidates[ 5 ] );
if ( cand_pen - candidates == 6 ) continue;
remove_factor( n, factor + candidates[ 6 ] );
}
cout << n;
done:
/*cout << endl;
for ( int hx = 0; hx < 8; ++ hx ) cout << (float) histo[hx] / total << " ";*/
cout << endl;
}
dkrauss$ /usr/local/bin/g++ main.cpp -o factorize_sse -O3 --profile-use
dkrauss$ time ./factorize_sse 18446743721522234449
4294967231 4294967279
real 0m13.437s
user 0m13.393s
sys 0m0.011s
以下是上述初稿。优化包括
remove_factor
。remove_factor
调用。remove_factor
调用完全展开最终循环,并确保该函数始终内联。
可读版本:
/*
* factorize_sse.cpp
* Filter out multiples of the first 17 primes while factorizing a number.
*
* Created by David Krauss on 10/14/10.
*
*/
#include <cmath>
#include <sstream>
#include <iostream>
#include <xmmintrin.h>
using namespace std;
int main( int argc, char *argv[] ) {
unsigned long n;
if ( argc != 2
|| ( istringstream( argv[1] ) >> n >> ws ).rdstate() != ios::eofbit ) {
cerr << "Usage: " << argv[0] << " <number>\n";
return 1;
}
int primes[] = { 2, 3, 5, 7, 11, 13, 17, 19, 23, 29, 31, 37, 41, 43, 47, 53, 59 };
for ( int *p = primes; p < primes + sizeof primes/sizeof *primes; ++ p ) {
while ( n % * p == 0 ) {
n /= * p;
cout << * p << " ";
}
}
if ( n != 1 ) {
__m128i mod = _mm_set_epi8( 1, 2, 3, 5, 6, 8, 9, 11, 14, 15, 18, 20, 21, 23, 26, 29 );
__m128i const prime = _mm_set_epi8( 3, 5, 7, 11, 13, 17, 19, 23, 29, 31, 37, 41, 43, 47, 53, 59 ),
one = _mm_set1_epi8( 1 );
for ( unsigned long factor = 1, limit = sqrtl( n ); factor < limit; ) {
factor += 2;
__m128i mask = _mm_cmpeq_epi8( mod, one ); // residual going to zero?
mod = _mm_sub_epi8( mod, one ); // update other residuals
if ( _mm_movemask_epi8( mask ) ) {
mask = _mm_and_si128( mask, prime ); // reset cycle if going to zero
mod = _mm_or_si128( mask, mod ); // combine reset into zeroed position
} else while ( n % factor == 0 ) {
n /= factor;
cout << factor << " ";
if ( n == 1 ) goto done;
}
}
cout << n;
}
done:
cout << endl;
}
答案 2 :(得分:5)
Fermat's factorization method简单快捷地找到成对的大素因子,只要你在它走得太远而变慢之前就停止它。然而,在我对随机数的测试中,这种情况太罕见,无法看到任何改进。
......没有使用概率方法(例如,Miller-Rabin)来确定素数
通过均匀分布,75%的输入将需要十亿次循环迭代,因此,即使您得到一个不确定的答案并且必须恢复到试验部门,也应该首先在不太确定的技术上花费一百万次操作。 / p>
我发现Pollard的Rho方法的布伦特变体非常好,但编码和理解更复杂。我见过的最好的例子是forum discussion。该方法依赖于运气,但经常有助于实现这一目标。
米勒 - 拉宾素性测试实际上是确定性的,大约10 ^ 15,这可以省去无果搜索的麻烦。
我尝试了几十种变体,并根据以下因素来确定int64值:
请注意,Pollard的Rho发现不一定是素数的因子,因此可以使用递归来计算因子。
答案 3 :(得分:2)
这段代码相当慢,我很确定我理解为什么。它并不是非常慢,但在10-20%的范围内肯定会更慢。每次循环都不应该进行一次除法,但唯一的方法就是调用sqrt
或类似的东西。
// system specific typedef for ulong should go here (or use boost::uint64_t)
typedef std::vector<ulong> ULongVector;
void GetFactors(ULongVector &factors, ulong num)
{
if (num<2)
return;
ulong workingNum=num;
ulong nextOffset=2;
while (workingNum%2==0)
{
factors.push_back(2);
workingNum/=2;
}
while (workingNum%3==0)
{
factors.push_back(3);
workingNum/=3;
}
ulong n = 5;
while ((workingNum != 1) && ((workingNum / n) >= n)) {
// Is workingNum divisible by n?
if (workingNum%n==0)
{
// n is a factor!
// so is the number multiplied by n to get workingNum
// Insert n into the list of factors
factors.push_back(n);
// Divide working number by n
workingNum/=n;
} else {
n+=nextOffset;
nextOffset=(nextOffset==2UL ? 4UL : 2UL);
}
}
if (workingNum != 1) {
// workingNum is prime, add it to the list of factors
factors.push_back(workingNum);
}
}
答案 4 :(得分:2)
结合Omnifarious的一些想法,以及其他改进:
// system specific typedef for ulong should go here (or use boost::uint64_t)
typedef unsigned __int64 ulong;
typedef std::vector<ulong> ULongVector;
// Caller needs to pass in an empty factors vector
void GetFactors(ULongVector &factors, ulong num)
{
if (num<2)
return;
ulong workingNum=num;
// Factor out factors of 2
while (workingNum%2==0)
{
factors.push_back(2);
workingNum/=2;
}
// Factor out factors of 3
while (workingNum%3==0)
{
factors.push_back(3);
workingNum/=3;
}
if (workingNum==1)
return;
// Factor out factors >=5
ulong nextOffset=2;
char nextShift = 1;
ulong n = 5;
ulong nn = 25;
do {
// Is workingNum divisible by n?
if (workingNum%n==0)
{
// n is a factor!
// so is the number multiplied by n to get workingNum
// Insert n into the list of factors
factors.push_back(n);
// Divide working number by n
workingNum/=n;
// Test for done...
if (workingNum==1)
return;
// Try n again
}
else {
nn += (n << (nextShift+1)) + (1<<(nextShift*2)); // (n+b)^2 = n^2 + 2*n*b + b*2
n += nextOffset;
nextOffset ^= 6;
nextShift ^= 3;
// invariant: nn == n*n
if (n & 0x100000000LL) break; // careful of integer wraparound in n^2
}
} while (nn <= workingNum);
// workingNum is prime, add it to the list of factors
factors.push_back(workingNum);
}
答案 5 :(得分:2)
自然的概括是使用比2和3更多的已知素数来预先计算跳过。像2,3,5,7,11一样,模式周期为2310(呵呵,数字不错)。也许更多,但它的收益递减 - 运行时间图可以确切地确定预计算开始产生负面影响的位置,但当然这取决于要考虑的数字的数量......
哈,我把编码细节留给你们。 : - )
干杯&amp;第h。,
- Alf
答案 6 :(得分:2)
我不确定这些会有多有效,但不是
while (workingNum%2==0)
你可以做到
while (workingNum & 1 == 0)
我不确定gcc或msvc(或者你使用的任何编译器)是否足够聪明来改变workingNum%2表达式,但是它很可能正在进行除法并查看edx中的模数...
我的下一个建议可能完全没有必要,具体取决于您的编译器,但您可以尝试在方法调用之前放置workingNum / = 3。 G ++可能足够聪明,可以看到不必要的除法,只需使用eax中的商(你也可以在更大的循环中执行此操作)。或者,更彻底(但更痛苦)的方法是内联汇编以下代码。
while (workingNum%3==0)
{
factors.push_back(3);
workingNum/=3;
}
编译器可能将模块化操作转换为divison,然后查看edx中的模数。问题是,你再次执行除法(我怀疑编译器是否看到你只是在循环条件下隐式执行除法)。所以,你可以内联汇编这个。这提出了两个问题:
1)方法调用push_back(3)。这可能会使寄存器混乱,完全没必要。
2)获取workingNum的注册表,但这可以通过初始模块化检查(将其强制为%eax)来确定,或者在当前时刻,它将/应该在eax中。
您可以将循环写为(假设workingNum在eax中,这是32位AT&amp; T语法,只是因为我不知道64位汇编或英特尔语法)
asm( "
movl $3, %ebx
WorkNumMod3Loop:
movl %eax, %ecx # just to be safe, backup workingNUm
movl $0, %edx # zero out edx
idivl $3 # divide by 3. quotient in eax, remainder in edx
cmpl $0, %edx # compare if it's 0
jne AfterNumMod3Loop # if 0 is the remainder, jump out
# no need to perform division because new workingNum is already in eax
#factors.push_back(3) call
je WorkNumMod3Loop
AfterNumMod3Loop:
movl %ecx, %eax"
);
您应该查看这些循环的程序集输出。您的编译器可能已经在进行这些优化,但我对此表示怀疑。如果在方法调用之前放置workingNum / = n可以在某些情况下稍微提高性能,我不会感到惊讶。