//MODIFIED VERSION: the earlier version wasn't compilable because i accidently left a comment 'uncommented'..Sorry for the inconvenience

// Take a string as input from the user. The string will contain only numbers
// ('0'-'9') and parentheses ( '(', ')' ). Your task is to find a section in the
// string that matches the pattern "k(q)", where k is a single digit, q is a
// substring of zero or more characters and the 2 parentheses form a matching set.
// Once such a pattern is found, replace it by 'k' occurrences of 'q'. Keep doing
// this till you can no longer find such patterns. Output the final string
// obtained.
// 
// Author:rahule@iitk.ac.in

#include<stdio.h>
int string_length(char *);	//function to find length of the string
void display(char *,int length);	
main()
{
  char s[50];
  
  printf("\nEnter the string\n");
  scanf("%s",s);
  //make sure the entered string is in the correct format.insert necessary checking procedures here,if needed.
  display(s,string_length(s));
  printf("\n");
}

void display(char *s,int length)
//reads the string pointed from *s upto length 'length' and   outputs in the correct format 
{
  int num,i,cnt,j;
  if(length==0)		//if length is 0,nothing is left to output.so return
     return;
  if(*(s+1)=='(')	//if a ''(' is encountered,find its pair ')', and calculate the length of the string between ( and ). Then recursively call the display function 
			//for the string between ( and )
  {
     num=(int)((*s)-'0');//since *(s+1)  is '(', *s should be the number indicating how many times the string between ( and ) has to be repeated.So,keep the integer
			//value of *s in num
     cnt=1;i=1;
     while(cnt!=0)	//finding the pair of '('
     {
	i++;
	if(s[i]=='(')	//whenever a new '(' is encountered,increment cnt
	cnt++;
	else if(s[i]==')')	//whenever a ')' is encountered decrement cnt
	     cnt--;
      }		//when cnt becomes 0,s[i] will be pointing to the ')' correspoinding to '(' encountered earlier
      
      for(j=0;j<num;j++)	//now 'display' the string between ( and ) ,'num' times..
      display(s+2,i-2);
      
      length=length-(i+1);	//since by above statement,we displayed the string between ( ),decremnt the length of the string from 'length'
      s=s+i+1;			//make s pointed to the character after the ')'
 
  if(length>0)			//if length is still +ve,continue with the next character
  {
    display(s,length);
  }
  else
    return;
 }
  
 else 
 {
       printf("%c",*s);
       display(++s,--length);
 }  
}

int string_length(char *s)
//funtion to find the string length
{
  int i;
  for(i=0;s[i]!='\0';i++);	//loop till the end of string
  return(i);
}