我已经将 问题＃12 从 欧拉计划 中选出作为编程练习，并比较了C，Python，Erlang和Haskell的实现（肯定不是最优的）。为了获得更高的执行时间，我搜索了第一个三角数，其除数超过1000个，而不是原问题中规定的500个。
结果如下： C：
结果如下： C：
lorenzo@enzo:~/erlang$ gcc -lm -o euler12.bin euler12.c
lorenzo@enzo:~/erlang$ time ./euler12.bin
842161320
real 0m11.074s
user 0m11.070s
sys 0m0.000s
Python：
lorenzo@enzo:~/erlang$ time ./euler12.py
842161320
real 1m16.632s
user 1m16.370s
sys 0m0.250s
使用PyPy的Python：
lorenzo@enzo:~/Downloads/pypy-c-jit-43780-b590cf6de419-linux64/bin$ time ./pypy /home/lorenzo/erlang/euler12.py
842161320
real 0m13.082s
user 0m13.050s
sys 0m0.020s
Erlang：
lorenzo@enzo:~/erlang$ erlc euler12.erl
lorenzo@enzo:~/erlang$ time erl -s euler12 solve
Erlang R13B03 (erts-5.7.4) [source] [64-bit] [smp:4:4] [rq:4] [async-threads:0] [hipe] [kernel-poll:false]
Eshell V5.7.4 (abort with ^G)
1> 842161320
real 0m48.259s
user 0m48.070s
sys 0m0.020s
Haskell：
lorenzo@enzo:~/erlang$ ghc euler12.hs -o euler12.hsx
[1 of 1] Compiling Main ( euler12.hs, euler12.o )
Linking euler12.hsx ...
lorenzo@enzo:~/erlang$ time ./euler12.hsx
842161320
real 2m37.326s
user 2m37.240s
sys 0m0.080s
总结:
- C: 100%
- Python: 692% (使用PyPy为118%)
- Erlang: 436% (由于RichardC的贡献为135%)
- Haskell: 1421%
我认为C有很大的优势，因为它使用长整型进行计算，而不是像其他三种语言一样使用任意长度的整数。此外，它不需要先加载运行时（其他语言需要吗？）。
问题1：
Erlang、Python和Haskell是否由于使用任意长度的整数而失去速度，或者只要值小于MAXINT就不会失去速度？
问题2： 为什么Haskell如此慢？有没有编译器标志可以关闭限制，或者是我的实现有问题？（后者很可能是因为Haskell对我来说是一本七个印章的书。）
问题3： 你能给我一些提示如何优化这些实现，而不改变我确定因子的方式吗？以任何方式进行优化：更好、更快、更符合语言的"本地"特性。
编辑:
问题4： 我的函数式实现是否允许LCO（尾递归消除），从而避免向调用堆栈添加不必要的帧？
我真的尽力在四种语言中尽可能相似地实现了相同的算法，尽管我必须承认我对Haskell和Erlang的了解非常有限。
使用的源代码：
#include <stdio.h>
#include <math.h>
int factorCount (long n)
{
double square = sqrt (n);
int isquare = (int) square;
int count = isquare == square ? -1 : 0;
long candidate;
for (candidate = 1; candidate <= isquare; candidate ++)
if (0 == n % candidate) count += 2;
return count;
}
int main ()
{
long triangle = 1;
int index = 1;
while (factorCount (triangle) < 1001)
{
index ++;
triangle += index;
}
printf ("%ld\n", triangle);
}
#! /usr/bin/env python3.2
import math
def factorCount (n):
square = math.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
triangle = 1
index = 1
while factorCount (triangle) < 1001:
index += 1
triangle += index
print (triangle)
-module (euler12).
-compile (export_all).
factorCount (Number) -> factorCount (Number, math:sqrt (Number), 1, 0).
factorCount (_, Sqrt, Candidate, Count) when Candidate > Sqrt -> Count;
factorCount (_, Sqrt, Candidate, Count) when Candidate == Sqrt -> Count + 1;
factorCount (Number, Sqrt, Candidate, Count) ->
case Number rem Candidate of
0 -> factorCount (Number, Sqrt, Candidate + 1, Count + 2);
_ -> factorCount (Number, Sqrt, Candidate + 1, Count)
end.
nextTriangle (Index, Triangle) ->
Count = factorCount (Triangle),
if
Count > 1000 -> Triangle;
true -> nextTriangle (Index + 1, Triangle + Index + 1)
end.
solve () ->
io:format ("~p~n", [nextTriangle (1, 1) ] ),
halt (0).
factorCount number = factorCount' number isquare 1 0 - (fromEnum $ square == fromIntegral isquare)
where square = sqrt $ fromIntegral number
isquare = floor square
factorCount' number sqrt candidate count
| fromIntegral candidate > sqrt = count
| number `mod` candidate == 0 = factorCount' number sqrt (candidate + 1) (count + 2)
| otherwise = factorCount' number sqrt (candidate + 1) count
nextTriangle index triangle
| factorCount triangle > 1000 = triangle
| otherwise = nextTriangle (index + 1) (triangle + index + 1)
main = print $ nextTriangle 1 1
Euler12[x_Integer] := Module[{s = 1}, For[i = 2, DivisorSigma[0, s] < x, i++, s += i]; s]。太棒了！ - tsvikas