The Java program below gives one possible answer in bold red.

class Parens {
   int level = 0;
   GetChar getChar = new GetChar();
   char next;
   String pOut = "";

   void P() {
      scan();
      Q();
      if (next == '#') System.out.println(pOut);
      else error(1);
   }

   void Q() {
      if (next == '(') {
         scan();
         pOut += "(" + level++ + " ";
      }
      else error(2);
      R();
   }

   void R() {
      if (next == ')') {
         scan();
         pOut += " " + --level + ")";
      }
      else {
         Q();
         R();
      }
   }

   void scan() {
      while (Character.isWhitespace(next = getChar.getNextChar()))
         ;
   }

   void error(int n) {
      System.out.println("*** ERROR: " + n);
      System.exit(1);
   }

   public static void main(String[] args) {
      Parens parens = new Parens();
      parens.P();
   }
}

Here are three separate runs (boldface = input):

(()())#
(0 (1  1)(1  1) 0)
( () (( )) (() () ) () ) #
(0 (1  1)(1 (2  2) 1)(1 (2  2)(2  2) 1)(1  1) 0)
((((((((((()))))))))))#
(0 (1 (2 (3 (4 (5 (6 (7 (8 (9 (10  10) 9) 8) 7) 6) 5) 4) 3) 2) 1) 0)
( () (( )) (()) () (((() (()) ) () ) )  )#
(0 (1  1)(1 (2  2) 1)(1 (2  2) 1)(1  1)(1 (2 (3 (4  4)(4 (5  5) 4) 3)(3  3) 2) 1) 0)