Erlang实现存在一些问题。作为下面的基准，我测量的未修改的Erlang程序的执行时间为47.6秒，而C代码的执行时间为12.7秒。

(编辑:在Erlang/OTP版本24,2021年，Erlang有一个自动JIT编译器，旧的+本机编译器选项不再支持或需要。我保留下面这段文字作为历史文件。关于export_all的注释对于jit生成良好代码的能力仍然是有效的。)

The first thing you should do if you want to run computationally intensive Erlang code is to use native code. Compiling with erlc +native euler12 got the time down to 41.3 seconds. This is however a much lower speedup (just 15%) than expected from native compilation on this kind of code, and the problem is your use of -compile(export_all). This is useful for experimentation, but the fact that all functions are potentially reachable from the outside causes the native compiler to be very conservative. (The normal BEAM emulator is not that much affected.) Replacing this declaration with -export([solve/0]). gives a much better speedup: 31.5 seconds (almost 35% from the baseline).

但是代码本身有一个问题:对于factorCount循环中的每一次迭代，都要执行以下测试:

factorCount (_, Sqrt, Candidate, Count) when Candidate == Sqrt -> Count + 1;

C代码不这样做。一般来说，在相同代码的不同实现之间进行公平的比较是很棘手的，特别是如果算法是数值的，因为您需要确保它们实际上在做相同的事情。在某个实现中由于某个类型转换而产生的轻微舍入错误可能会导致它比另一个实现进行更多的迭代，即使两者最终得到相同的结果。

为了消除这个可能的错误源(并在每次迭代中摆脱额外的测试)，我重写了factorCount函数，如下所示，密切模仿C代码:

factorCount (N) ->
    Sqrt = math:sqrt (N),
    ISqrt = trunc(Sqrt),
    if ISqrt == Sqrt -> factorCount (N, ISqrt, 1, -1);
       true          -> factorCount (N, ISqrt, 1, 0)
    end.

factorCount (_N, ISqrt, Candidate, Count) when Candidate > ISqrt -> Count;
factorCount ( N, ISqrt, Candidate, Count) ->
    case N rem Candidate of
        0 -> factorCount (N, ISqrt, Candidate + 1, Count + 2);
        _ -> factorCount (N, ISqrt, Candidate + 1, Count)
    end.

这个重写，没有export_all和本机编译，给了我以下运行时:

$ erlc +native euler12.erl
$ time erl -noshell -s euler12 solve
842161320

real    0m19.468s
user    0m19.450s
sys 0m0.010s

这与C代码相比不算太糟:

$ time ./a.out 
842161320

real    0m12.755s
user    0m12.730s
sys 0m0.020s

考虑到Erlang完全不适合编写数字代码，在这样的程序中只比C慢50%就已经很不错了。

最后，关于你的问题:

问题1:erlang、python和haskell是否会因为使用任意长度的整数而降低速度 只要值小于MAXINT，它们不就行了吗?

Yes, somewhat. In Erlang, there is no way of saying "use 32/64-bit arithmetic with wrap-around", so unless the compiler can prove some bounds on your integers (and it usually can't), it must check all computations to see if they can fit in a single tagged word or if it has to turn them into heap-allocated bignums. Even if no bignums are ever used in practice at runtime, these checks will have to be performed. On the other hand, that means you know that the algorithm will never fail because of an unexpected integer wraparound if you suddenly give it larger inputs than before.

问题4:我的函数实现是否允许LCO，从而避免在调用堆栈中添加不必要的帧?

是的，您的Erlang代码在最后调用优化方面是正确的。

看看您的Erlang实现。计时包括启动整个虚拟机、运行程序和停止虚拟机。我很确定设置和停止erlang vm需要一些时间。

If the timing was done within the erlang virtual machine itself, results would be different as in that case we would have the actual time for only the program in question. Otherwise, i believe that the total time taken by the process of starting and loading of the Erlang Vm plus that of halting it (as you put it in your program) are all included in the total time which the method you are using to time the program is outputting. Consider using the erlang timing itself which we use when we want to time our programs within the virtual machine itself timer:tc/1 or timer:tc/2 or timer:tc/3. In this way, the results from erlang will exclude the time taken to start and stop/kill/halt the virtual machine. That is my reasoning there, think about it, and then try your bench mark again.

实际上，我建议我们尝试在这些语言的运行时内为程序计时(对于具有运行时的语言)，以便获得精确的值。例如，C不像Erlang、Python和Haskell那样有启动和关闭运行时系统的开销(98%确定-我可以纠正)。因此(基于这个推理)我总结说，这个基准测试对于运行在运行时系统之上的语言来说不够精确/公平。让我们用这些更改再做一次。

编辑:此外，即使所有的语言都有运行时系统，启动和停止它们的开销也会有所不同。因此，我建议我们从运行时系统内部计时(对于应用此方法的语言)。众所周知，Erlang VM在启动时有相当大的开销!

问题3:你能给我一些如何优化这些实现的提示吗 而不改变我确定因子的方法?任意优化 方法:更好、更快、更“地道”的语言。

C实现是次优的(正如Thomas M. DuBuisson所暗示的那样)，该版本使用64位整数(即长数据类型)。稍后我将研究程序集清单，但根据合理的猜测，在编译后的代码中进行了一些内存访问，这使得使用64位整数明显变慢。或者是生成的代码(比如在SSE寄存器中可以容纳更少的64位整数，或者将双精度整数舍入为64位整数更慢)。

下面是修改后的代码(简单地用int替换long，我显式内联factorCount，尽管我不认为这是gcc -O3所必需的):

#include <stdio.h>
#include <math.h>

static inline int factorCount(int n)
{
    double square = sqrt (n);
    int isquare = (int)square;
    int count = isquare == square ? -1 : 0;
    int candidate;
    for (candidate = 1; candidate <= isquare; candidate ++)
        if (0 == n % candidate) count += 2;
    return count;
}

int main ()
{
    int triangle = 1;
    int index = 1;
    while (factorCount (triangle) < 1001)
    {
        index++;
        triangle += index;
    }
    printf ("%d\n", triangle);
}

运行+计时它给出:

$ gcc -O3 -lm -o euler12 euler12.c; time ./euler12
842161320
./euler12  2.95s user 0.00s system 99% cpu 2.956 total

作为参考，Thomas在前面的回答中给出了haskell实现:

$ ghc -O2 -fllvm -fforce-recomp euler12.hs; time ./euler12                                                                                      [9:40]
[1 of 1] Compiling Main             ( euler12.hs, euler12.o )
Linking euler12 ...
842161320
./euler12  9.43s user 0.13s system 99% cpu 9.602 total

结论:ghc是一个很棒的编译器，但gcc通常会生成更快的代码。

在Python优化方面，除了使用PyPy(对代码进行零更改即可获得令人印象深刻的加速)之外，还可以使用PyPy的翻译工具链编译与rpython兼容的版本，或者使用Cython构建扩展模块，在我的测试中，这两种工具都比C版本快，而Cython模块的速度几乎是C版本的两倍。作为参考，我包括C和PyPy基准测试结果:

C(编译gcc -O3 -lm)

% time ./euler12-c 
842161320

./euler12-c  11.95s 
 user 0.00s 
 system 99% 
 cpu 11.959 total

PyPy 1.5

% time pypy euler12.py
842161320
pypy euler12.py  
16.44s user 
0.01s system 
99% cpu 16.449 total

RPython(使用最新的PyPy修订版，c2f583445aee)

% time ./euler12-rpython-c
842161320
./euler12-rpy-c  
10.54s user 0.00s 
system 99% 
cpu 10.540 total

崇拜0.15

% time python euler12-cython.py
842161320
python euler12-cython.py  
6.27s user 0.00s 
system 99% 
cpu 6.274 total

RPython版本有几个关键的变化。要转换成一个独立的程序，您需要定义目标，在本例中是主函数。它被期望接受sys。Argv作为它唯一的参数，并且需要返回一个int。你可以使用translate.py， % translate.py euler12-rpython.py来翻译它，它可以翻译成C语言并为你编译它。

# euler12-rpython.py

import math, sys

def factorCount(n):
    square = math.sqrt(n)
    isquare = int(square)
    count = -1 if isquare == square else 0
    for candidate in xrange(1, isquare + 1):
        if not n % candidate: count += 2
    return count

def main(argv):
    triangle = 1
    index = 1
    while factorCount(triangle) < 1001:
        index += 1
        triangle += index
    print triangle
    return 0

if __name__ == '__main__':
    main(sys.argv)

def target(*args):
    return main, None

Cython版本被重写为扩展模块_euler12。我从一个普通的python文件中导入并调用它。_euler12。Pyx本质上与您的版本相同，只是有一些额外的静态类型声明。setup.py有一个正常的样板来构建扩展，使用python setup.py build_ext——inplace。

# _euler12.pyx
from libc.math cimport sqrt

cdef int factorCount(int n):
    cdef int candidate, isquare, count
    cdef double square
    square = sqrt(n)
    isquare = int(square)
    count = -1 if isquare == square else 0
    for candidate in range(1, isquare + 1):
        if not n % candidate: count += 2
    return count

cpdef main():
    cdef int triangle = 1, index = 1
    while factorCount(triangle) < 1001:
        index += 1
        triangle += index
    print triangle

# euler12-cython.py
import _euler12
_euler12.main()

# setup.py
from distutils.core import setup
from distutils.extension import Extension
from Cython.Distutils import build_ext

ext_modules = [Extension("_euler12", ["_euler12.pyx"])]

setup(
  name = 'Euler12-Cython',
  cmdclass = {'build_ext': build_ext},
  ext_modules = ext_modules
)

老实说，我对RPython或Cython都没有什么经验，对结果感到惊喜。如果您正在使用CPython，那么在Cython扩展模块中编写cpu密集型代码似乎是优化程序的一种非常简单的方法。

使用Haskell，您真的不需要显式地考虑递归。

factorCount number = foldr factorCount' 0 [1..isquare] -
                     (fromEnum $ square == fromIntegral isquare)
    where
      square = sqrt $ fromIntegral number
      isquare = floor square
      factorCount' candidate
        | number `rem` candidate == 0 = (2 +)
        | otherwise = id

triangles :: [Int]
triangles = scanl1 (+) [1,2..]

main = print . head $ dropWhile ((< 1001) . factorCount) triangles

在上面的代码中，我用普通的列表操作替换了@Thomas回答中的显式递归。代码仍然做着完全相同的事情，而不需要我们担心尾部递归。它运行(~ 7.49秒)比@Thomas回答的版本(~ 7.04秒)在我的机器上运行GHC 7.6.2，而来自@Raedwulf的C版本运行~ 3.15秒。GHC似乎在过去一年中有所改善。

PS:我知道这是一个老问题，我从谷歌搜索中偶然发现了它(我忘了我在搜索什么了，现在…)只是想评论一下关于LCO的问题，并表达我对Haskell的总体感受。我想对上面的答案进行注释，但是注释不允许代码块。

只是为了好玩。下面是一个更“原生”的Haskell实现:

import Control.Applicative
import Control.Monad
import Data.Either
import Math.NumberTheory.Powers.Squares

isInt :: RealFrac c => c -> Bool
isInt = (==) <$> id <*> fromInteger . round

intSqrt :: (Integral a) => a -> Int
--intSqrt = fromIntegral . floor . sqrt . fromIntegral
intSqrt = fromIntegral . integerSquareRoot'

factorize :: Int -> [Int]
factorize 1 = []
factorize n = first : factorize (quot n first)
  where first = (!! 0) $ [a | a <- [2..intSqrt n], rem n a == 0] ++ [n]

factorize2 :: Int -> [(Int,Int)]
factorize2 = foldl (\ls@((val,freq):xs) y -> if val == y then (val,freq+1):xs else (y,1):ls) [(0,0)] . factorize

numDivisors :: Int -> Int
numDivisors = foldl (\acc (_,y) -> acc * (y+1)) 1 <$> factorize2

nextTriangleNumber :: (Int,Int) -> (Int,Int)
nextTriangleNumber (n,acc) = (n+1,acc+n+1)

forward :: Int -> (Int, Int) -> Either (Int, Int) (Int, Int)
forward k val@(n,acc) = if numDivisors acc > k then Left val else Right (nextTriangleNumber val)

problem12 :: Int -> (Int, Int)
problem12 n = (!!0) . lefts . scanl (>>=) (forward n (1,1)) . repeat . forward $ n

main = do
  let (n,val) = problem12 1000
  print val

使用ghc -O3，它在我的机器上持续运行0.55-0.58秒(1.73GHz Core i7)。

C版本中一个更有效的factorCount函数:

int factorCount (int n)
{
  int count = 1;
  int candidate,tmpCount;
  while (n % 2 == 0) {
    count++;
    n /= 2;
  }
    for (candidate = 3; candidate < n && candidate * candidate < n; candidate += 2)
    if (n % candidate == 0) {
      tmpCount = 1;
      do {
        tmpCount++;
        n /= candidate;
      } while (n % candidate == 0);
       count*=tmpCount;
      }
  if (n > 1)
    count *= 2;
  return count;
}

在main中使用gcc -O3 -lm将long类型更改为int类型，该程序始终在0.31-0.35秒内运行。

如果您利用第n个三角形数= n*(n+1)/2，并且n和(n+1)具有完全不同的质因数分解，则可以使两者运行得更快，因此可以将每个一半的因数数相乘，以得到整体的因数数。以下几点:

int main ()
{
  int triangle = 0,count1,count2 = 1;
  do {
    count1 = count2;
    count2 = ++triangle % 2 == 0 ? factorCount(triangle+1) : factorCount((triangle+1)/2);
  } while (count1*count2 < 1001);
  printf ("%lld\n", ((long long)triangle)*(triangle+1)/2);
}

将c代码的运行时间减少到0.17-0.19秒，它可以处理更大的搜索——大于10000个因数在我的机器上大约需要43秒。我给感兴趣的读者留下了类似的haskell加速。