刚刚学习hash，做了两道题，挺有体会的。还在摸索中，代码写的很乱……
http://acm.pku.edu.cn/JudgeOnline/problem?id=2549
// 首先枚举a+b，建立hashtable，（注意a+b小于0时的取址问题），
 // 然后枚举d-c，同样注意取址，在hashtable中查找地址相同的，
 // 并比较是否是所查的结果.记录当前最大的d值 
 // 注意，i，j，k，l不能重复…… 
#include < iostream >
#include < cmath >
using   namespace  std;
 int  n;
 struct  q
{    long  val;
     int  x1,x2;
}hash[ 1000001 ];
 int  prime = 999983 ;
 long  num[ 10001 ]; 
 long  sum,cha;
 int  i,j;
 void  build( long  t)
{    if (hash[t].val == 0 )
    {  hash[t].val = sum;
       hash[t].x1 = i;
       hash[t].x2 = j;
    }
     else  
    {    if (hash[t].val != 0 && hash[t].val != sum)
            build(t + 1 );
    }
}
 bool  find( long  t)
{    if (hash[t].val == 0 )
         return   0 ;
     else   if (hash[t].val == cha && hash[t].x1 != i && hash[t].x2 != i && hash[t].x1 != j && hash[t].x2 != j)
        return   1 ;
     else   if (hash[t].val != 0 && hash[t].val != cha)
    {   find(t + 1 );
    }
     else   return   0 ;
}
 int  main()
{    while (scanf( " %d " , & n) && n)
    {   memset(hash, 0 , sizeof (hash));
        memset(num, 0 , sizeof (num));
         long  k;
         long  temp;
         for (i = 0 ;i < n;i ++ )
        {   scanf( " %ld " , & num[i]);
        }
         for (i = 0 ;i < n;i ++ )
        {    for (j = 0 ;j < n;j ++ )
            {     if (i == j)
                      continue ;
                 sum = num[i] + num[j];
                  if (sum < 0 )
                    temp = ( - 1 ) * sum;
                  else  temp = sum;
                 build(temp % prime);
            }
        } // build hashtable[]=a+b;
         long  best =- 536870913 ;
         for (i = 0 ;i < n;i ++ )
        {   for (j = 0 ;j < n;j ++ )
           {    if (i == j)
                  continue ;
               cha = num[i] - num[j];
                if (cha < 0 )
                  temp = ( - 1 ) * cha;
                else  temp = cha;
                if (find(temp % prime) && num[i] > best)
                   best = num[i];
           }
        } // 穷举hash[x]=d-c==a+b 
         if (best ==- 536870913 )
            printf( " no solution\n " );
         else
            printf( " %ld\n " ,best);
    }
    system( " pause " );
     return   0 ;
}