可笑的优化

    这几天没事做的时候都会上projecteuler.net上面去做题，其中14题是这样的：
he following iterative sequence is defined for the set of positive integers:

n n/2 (n is even)
n 3n + 1 (n is odd)

Using the rule above and starting with 13, we generate the following sequence:

13 40 20 10 5 16 8 4 2 1

It can be seen that this sequence (starting at 13 and finishing at 1) contains 10 terms. Although it has not been proved yet (Collatz Problem), it is thought that all starting numbers finish at 1.

Which starting number, under one million, produces the longest chain?

    题目并不难理解，这个据说是著名的角谷猜想，现在要找到100万以下的数字中展开这个链最长的数字是多少。如果我一开始就直接按照题意来解答，这个题目花不了几分钟，直接暴力法。然而我却想的太多了，我猜想在计算这个链条长度的过程中会不会有很多数字会重复计算，如果加上缓存以前计算的结果是否能节约比较多的时间？那么第一次解答如下：

#include < iostream >
#include < map >
#include < windows.h >
using   namespace  std;
unsigned  long  produce_term(unsigned  long  n)
{
     if (n & 1 )
         return   3 * n + 1 ;
     else
         return  n >> 1 ;
}
int  main()
{
    map < unsigned  long , int >  counters;
     int  max_i = 0 ;
     int  max_count = 0 ;
    DWORD tick1,tickPassed;
    tick1  =  GetTickCount(); 
     for ( int  i = 1 ;i < 1000000 ;i ++ )
    {
         int  sum = 2 ;
        unsigned  long  term = i;
         while ((term = produce_term(term)) != 1 )
        {
             if (counters[term]){
                sum += counters[term];
                 break ;
            } else
                sum += 1 ;
        }

         if (sum > max_count)
        {
            max_i = i;
            max_count = sum;
            counters[i] = sum;
        }

    }
    tickPassed  =  GetTickCount() - tick1; 
    cout << tickPassed << endl;
    cout << max_i << endl << max_count << endl;
     return   0 ;
}

  
    遗憾的是，这个版本跑了快13分钟，太让人难以接受了。那么是否能优化下？怎么优化？我的机器是双核的，跑这个单进程单线程的程序只利用了一半的CPU，那么能不能搞成 两个线程来计算？缓存需要在两个线程之间做同步，显然读的多，写的少，应该采用 读写锁。OK，第二个版本利用ACE的线程封装实现如下：

#include < iostream >
#include < map >
#include  " ace/Thread_mutex.h "
#include  " ace/Synch.h "
#include  " ace/Thread_Manager.h "
using   namespace  std;
class  ThreadSafeMap
{
public :
    ThreadSafeMap()
    {
    }
     int   get (unsigned  long  n)
    {
        ACE_READ_GUARD_RETURN(ACE_RW_Thread_Mutex,guard,mutex_, 0 );
         return  counters_[n];
    }
     int  put(unsigned  long  key, int  value)
    {
        ACE_WRITE_GUARD_RETURN(ACE_RW_Thread_Mutex,guard,mutex_, - 1 );
        counters_[key] = value;
         return   0 ;
    }

private :
    map < unsigned  long , int >  counters_;
    ACE_RW_Thread_Mutex mutex_;
};
unsigned  long  produce_term(unsigned  long  n)
{
     if (n & 1 )
         return   3 * n + 1 ;
     else
         return  n >> 1 ;
}
static  ThreadSafeMap counters;
ACE_THR_FUNC_RETURN run_svc ( void   * arg)
{
     int  max_i = 0 ;
     int  max_count = 0 ;
     for ( int  i = 500001 ;i < 1000000 ;i ++ )
    {
         int  sum = 2 ;
        unsigned  long  term = i;
         while ((term = produce_term(term)) != 1 )
        {
             if (counters. get (term)){
                sum += counters. get (term);
                 break ;
            } else
                sum += 1 ;
        }

         if (sum > max_count)
        {
            max_i = i;
            max_count = sum;
            counters.put(i,sum);
        }

    }
    cout << max_i << endl << max_count << endl;
     return   0 ;
}
int  main( int  ac, char *  argv[])
{
     if  (ACE_Thread_Manager::instance () -> spawn (
         //  Pointer to function entry point.
        run_svc,
         //  <run_svc> parameter.
        NULL,
        THR_DETACHED  |  THR_SCOPE_SYSTEM)  ==   - 1 )
         return   - 1 ;
     int  max_i = 0 ;
     int  max_count = 0 ;

     for ( int  i = 1 ;i < 500000 ;i ++ )
    {
         int  sum = 2 ;
        unsigned  long  term = i;
         while ((term = produce_term(term)) != 1 )
        {
             if (counters. get (term)){
                sum += counters. get (term);
                 break ;
            } else
                sum += 1 ;
        }

         if (sum > max_count)
        {
            max_i = i;
            max_count = sum;
            counters.put(i,sum);
        }

    }
    cout << max_i << endl << max_count << endl;
     return  ACE_Thread_Manager::instance () -> wait ();
}

   
    将数据分成了两半，利用两个线程来计算，果然快了一点，快了多少呢？从13分钟减少到9分钟，CPU利用率也到了100%，内存占用也降低了一半，似乎成绩不错呀。正在沾沾自喜之际，突然想起，能不能简单地暴力破解，咱不搞缓存，不搞多线程，看看效果怎么样。那么第三个版本简单实现如下：

#include < iostream >
using   namespace  std;
unsigned  long  produce_term(unsigned  long  n)
{
     if (n & 1 )
         return   3 * n + 1 ;
     else
         return  n >> 1 ;
}
int  main()
{
   int  max_i;
   int  max_count = 0 ;
   for ( int  i = 1 ;i < 1000000 ;i ++ )
  {
      int  count = 2 ;
     unsigned  long  term = i;
      while ((term = produce_term(term)) > 1 )
         count += 1 ;
      if (count > max_count){
           max_i = i;
           max_count = count;
     }
  }
  cout << max_i << endl << max_count << endl;
  system( " pause " );
   return   0 ;
}

    程序执行的结果让我惊掉了下巴，竟然只执行了1秒多，换成java也是一样。什么缓存、多线程，全抛到了九霄云外。

     总结教训， 想当然的性能估计是愚不可及的，想当然的优化是愚不可及的，简单直接才是美！