耽搁了很久的题目，突然想起来了这题还没做，这几天做了几道字典树，就拿来用字典树做一做。速度不是很快，空间用的很大。至少ac了……
#include < iostream >
#include < cstring >
using   namespace  std;
 int  n;
 char  tmp[ 100 ]; // 输入字符串 
struct  list
{
     int  val; // 7位数字的值 
     int  num[ 8 ]; // 7位数字 
}s[ 100001 ]; // 输入的list 
struct  ans
{
     int  val;
     int  num[ 8 ];
}xx[ 101 ]; // 满足cnt>=2的list 
struct  Trie
{
     int  next[ 10 ];
     int  cnt;
     void  init(){memset(next, - 1 , sizeof (next));cnt = 0 ;}
};
Trie T[ 200001 ];
 int  p,q = 0 ;
 int  cmp( const   void   * a, const   void   * b) // 按val值排序 
{
     struct  ans  * c = (ans * )a;
     struct  ans  * d = (ans * )b;
     return  c -> val > d -> val ? 1 : - 1 ;
}
 int  turn( char  c[], int  t) // 将字符串转换为数字 
{
     int  i,j;
    j = 0 ;
     for (i = 0 ;i < strlen(c);i ++ )
    {
         if (c[i] >= ' A ' && c[i] <= ' C ' )
        {
            s[t].num[j ++ ] = 2 ;
        }
         else   if (c[i] >= ' D ' && c[i] <= ' F ' )
        {
            s[t].num[j ++ ] = 3 ;
        }
         else   if (c[i] >= ' G ' && c[i] <= ' I ' )
        {
            s[t].num[j ++ ] = 4 ;
        }
         else   if (c[i] >= ' J ' && c[i] <= ' L ' )
        {
            s[t].num[j ++ ] = 5 ;
        }
         else   if (c[i] >= ' M ' && c[i] <= ' O ' )
        {
            s[t].num[j ++ ] = 6 ;
        }
         else   if (c[i] >= ' P ' && c[i] <= ' S ' )
        {
            s[t].num[j ++ ] = 7 ;
        }
         else   if (c[i] >= ' T ' && c[i] <= ' V ' )
        {
            s[t].num[j ++ ] = 8 ;
        }
         else   if (c[i] >= ' W ' && c[i] <= ' Y ' )
        {
            s[t].num[j ++ ] = 9 ;
        }
         else   if (c[i] >= ' 0 ' && c[i] <= ' 9 ' )
        {
            s[t].num[j ++ ] = c[i] - ' 0 ' ;
        }
    }
    s[t].val = s[t].num[ 0 ] * 1000000 + s[t].num[ 1 ] * 100000 + s[t].num[ 2 ] * 10000 + s[t].num[ 3 ] * 1000 + s[t].num[ 4 ] * 100 + s[t].num[ 5 ] * 10 + s[t].num[ 6 ];
}
 int  build() // 建树 
{
    p = 1 ;
    T[p].init();
     return   0 ;
}
 int  insert( int  c[], int  t) // 插入电话号码 
{
     int  indx = 1 ;
     int  i,j;
     for (i = 0 ;i < 7 ;i ++ )
    {
         if (T[indx].next[c[i]] ==- 1 )
        {
            T[indx].next[c[i]] =++ p;
            T[p].init();
        }
        indx = T[indx].next[c[i]];
    }
    T[indx].cnt ++ ;
     if (T[indx].cnt == 2 )
    {
        memcpy(xx[q].num,s[t].num, sizeof (s[t].num));
        xx[q ++ ].val = s[t].val;
    }
}
 int  check( int  c[]) // 查找>2的电话条目的cnt 
{
     int  i;
     int  indx = 1 ;
     for (i = 0 ;i < 7 ;i ++ )
    {
        indx = T[indx].next[c[i]];
    }
     return  T[indx].cnt;
}
 int  main()
{
     int  n;
     int  i,j;
    scanf( " %d " , & n);
    build();
     for (i = 1 ;i <= n;i ++ )
    {
        scanf( " %s " ,tmp);
        turn(tmp,i); 
        insert(s[i].num,i);
    }
     if (q == 0 )
    {
        printf( " No duplicates.\n " );
         return   0 ;
    }
     else
    {
        qsort(xx,q, sizeof (xx[ 0 ]),cmp); 
        for (i = 0 ;i < q;i ++ )
        {
            for (j = 0 ;j < 3 ;j ++ )
            {
                printf( " %d " ,xx[i].num[j]);
            }
            printf( " - " );
            for (;j < 7 ;j ++ )
            {
                printf( " %d " ,xx[i].num[j]);
            }
            printf( "   " );
            printf( " %d\n " ,check(xx[i].num));
        }
    }
    system( " pause " );
     return   0 ;
}