UVa 389 Basically Speaking

又一道基础题，进制转换。

做水题也不能大意！

  
    

   
     
   
      1 
   
     //
   
     68ms 2010-05-14 15:48:58

   
      2 
   
      
   
     //
   
     log: wa(1) ac(1)

   
      3 
   
     
   
     //
   
     Type: Base number conversion(进制转换)

   
      4 
   
     
   
     //
   
     key1: strrev() 在ANSI C, C++ 中不能使用

   
      5 
   
     
   
     //
   
     key2: 0的时候要处理

   
      6 
   
     
   
     //
   
     Thinking: After finished coding, member to try some special data.
   
     

   
      7 
   
     
   
     #include 
   
     <
   
     stdio.h
   
     >
   
     

   
      8 
   
     #include 
   
     <
   
     string
   
     .h
   
     >
   
     

   
      9 
   
     
   
     #define
   
      NL 100
   
     

   
     10 
   
     

   
     11 
   
     
   
     int
   
      
   
     get
   
     (
   
     char
   
      a)

   
     12 
   
     {

   
     13 
   
      
   
     if
   
      (a 
   
     >=
   
      
   
     '
   
     0
   
     '
   
      
   
     &&
   
      a 
   
     <=
   
      
   
     '
   
     9
   
     '
   
     ) 
   
     return
   
      a
   
     -
   
     '
   
     0
   
     '
   
     ;

   
     14 
   
      
   
     return
   
      a
   
     -
   
     '
   
     A
   
     '
   
     +
   
     10
   
     ;

   
     15 
   
     }

   
     16 
   
     

   
     17 
   
     
   
     int
   
      main()

   
     18 
   
     {

   
     19 
   
      
   
     char
   
      s0[NL], s1[NL];

   
     20 
   
      
   
     int
   
      len, I, K, M, b1, b2;

   
     21 
   
      
   
     while
   
      (scanf(
   
     "
   
     %s%d%d
   
     "
   
     , s0, 
   
     &
   
     b1, 
   
     &
   
     b2) 
   
     !=
   
      EOF) {

   
     22 
   
      len 
   
     =
   
      strlen(s0);

   
     23 
   
      
   
     int
   
      t 
   
     =
   
      
   
     0
   
     , bb 
   
     =
   
      
   
     1
   
     ;

   
     24 
   
      I 
   
     =
   
      len 
   
     -
   
      
   
     1
   
     ;

   
     25 
   
      
   
     while
   
      (I
   
     >=
   
     0
   
     ) {

   
     26 
   
      t 
   
     +=
   
      
   
     get
   
     (s0[I])
   
     *
   
     bb;

   
     27 
   
      bb 
   
     *=
   
      b1;

   
     28 
   
      I
   
     --
   
     ;

   
     29 
   
      }

   
     30 
   
      K 
   
     =
   
      
   
     0
   
     ;

   
     31 
   
      
   
     while
   
      (t
   
     >
   
     0
   
     ) {

   
     32 
   
      I 
   
     =
   
      t
   
     %
   
     b2;

   
     33 
   
      
   
     if
   
      (I
   
     >=
   
     10
   
     ) s1[K
   
     ++
   
     ] 
   
     =
   
      I
   
     -
   
     10
   
     +
   
     '
   
     A
   
     '
   
     ;

   
     34 
   
      
   
     else
   
      s1[K
   
     ++
   
     ] 
   
     =
   
      I
   
     +
   
     '
   
     0
   
     '
   
     ;

   
     35 
   
      t 
   
     /=
   
      b2;

   
     36 
   
      }

   
     37 
   
      
   
     if
   
      (K 
   
     ==
   
      
   
     0
   
     ) { s1[K
   
     ++
   
     ] 
   
     =
   
      
   
     '
   
     0
   
     '
   
     ; }

   
     38 
   
      s1[K] 
   
     =
   
      
   
     '
   
     \0
   
     '
   
     ;

   
     39 
   
      
   
     if
   
      (K 
   
     >
   
      
   
     7
   
     ) puts(
   
     "
   
      ERROR
   
     "
   
     );

   
     40 
   
      
   
     else
   
      {

   
     41 
   
      
   
     for
   
      (I
   
     =
   
     0
   
     , M
   
     =
   
     K
   
     -
   
     1
   
     ; M
   
     >=
   
     0
   
     ; M
   
     --
   
     ) {

   
     42 
   
      s0[I
   
     ++
   
     ] 
   
     =
   
      s1[M];

   
     43 
   
      }

   
     44 
   
      s0[I] 
   
     =
   
      
   
     '
   
     \0
   
     '
   
     ;

   
     45 
   
      printf(
   
     "
   
     %7s\n
   
     "
   
     , s0);

   
     46 
   
      }

   
     47 
   
      }

   
     48 
   
      
   
     return
   
      
   
     0
   
     ;

   
     49 
   
     }

   
     50 
   
     
   
     /*
   
     

   
     51 
   
     input:

   
     52 
   
     0000000 2 2

   
     53 
   
     
   
     */