Google China Code jam round2 div1 1000 question

I did not open the question until the end to see what the 1000 question is.
Google also played a psychological game.
The simplest fraction is used to represent the equivalent repeating decimal places. The denominator must be a positive integer.
Enter a string of 3 to 10 characters (inclusive): A. B (c) to indicate a circular decimal number.
Enter the following conditions to ensure:
The character is only 0123456789 characters .()
A Indicates the integer part. A maximum of 1 digits (2nd characters are always '.')
B Indicates the non-repeating part, which can be at least 0 bits
C Indicates the cyclic part, which can be at least 0 bits.
Parentheses must appear in pairs or not.
Returns the score of a/B.

Test data:
"0. (3 )"
Returns: "1/3"
"1.3125"
Returns: "21/16"
"2.85 (23 )"
Returns: "14119/4950"
9.123 (456 )"
Returns: "3038111/333000"
0.111 (1 )"
Returns: "1/9"
"3. (000 )"
Returns: "3/1"

Below is what I wrote laterCode.

# Include < String >
# Include < Cstdio >
Using   Namespace STD;
Class Recurringnumbers
{
Public :
Int Gcd ( Int A, Int B)
{
Int T;
If ( < B) T = A, = B, B = T;
While (B)
{
T=A%B;
A=B;
B=T;
}
Return A;
}
String Converttofraction ( Const   String   & Decimalnumber)
{
Char Buf [ 32 ], [ 10 ], B [ 10 ];
Int Lena;
Strcpy (BUF, decimalnumber. c_str ());
Strcpy (A, strtok (BUF, " .() " ));
Char   * N = Strtok (null, " .() " );
Char   * N2 = Strtok (null, " .() " );
If (N2 ! = Null)
{
Strcat (A, N );
Strcpy (B, N2 );
Lena=Strlen (N );
}
Else
{
If (Strchr (decimalnumber. c_str (), ' ( ' ))
{
Lena=0;
Strcpy (B, n );
}
Else
{
Lena=Strlen (N );
Strcat (A, N );
Strcpy (B,"0");
}
}
Int Na = Atoi ();
Int NB = Atoi (B );
Int Nd = 1 ;
Int I, Ng, NT;
For (NT = 1 , I = 0 ; I < Lena; I ++ )
NT = NT * 10 ;
If (NB ! =   0 )
{
For (Nd = I = 0 ; B [I]; I ++ )
Nd = Nd * 10 + 9 ;
Na = Na * Nd + NB;
NT * = Nd;
}
NG = Gcd (Na, NT );
Na /= Ng;
NT /= Ng;
Sprintf (BUF, " % D/% d " , Na, NT );
Return   String (BUF );
}
} ;
# If 0
Int Main ()
{
Char   * S [] =
{
" 0. (0) " , // Returns: "1/3"
" 1.3125 " , // Returns: "21/16"
" 2.85 (23) " , // Returns: "14119/4950"
" 9.123 (456) " , // Returns: "3038111/333000"
" 0.111 (1) " , // Returns: "1/9"
" 3. (000) " , // Returns: "3/1"
0 ,
} ;
Recurringnumbers;
For ( Int I = 0 ; S [I]; I ++ )
Printf ( " % S \ n " , A. converttofraction ( String (S [I]). c_str ());

Return   0 ;
}
# Endif