// For a character array consisting of 0, 1 and 'x', find 101 numbers in the array.
// A number is a 101 number if there exists some i and j such that the ith
// character is 1, the (i+j)th character is 0 and the (i+2j)th character is 1.

//Author:rahule@cse.iitk.ac.in

#include<stdio.h>
int string_length(char *);	//function to find the string length
main()
{
  char a[50];
  int i,j,n;
  printf("\nEnter the array elements (0,1 or x) :\n");
  scanf("%s",a);
  n=string_length(a);	//calculating the string lenghth entered
  
  for(i=0;i<n;i++)
    for(j=i+1;2*j<n;j++)
    {
      if((a[i]=='1' || a[i]=='x')&&(a[i+j]=='0' || a[i+j]=='x')&&(a[i+2*j]=='1' || a[i+2*j]=='x'))	//checking the condition of 101 number
      {
	printf("\n%c%c%c ",a[i],a[i+j],a[i+2*j]);	//prining the number
	
	if(a[i]=='x' || a[i+j]=='x' || a[i+2*j]=='x')	//printing to which number x is treated as(if present)
	{
	  printf("(");
	  if(a[i]=='x')
	    printf("x as 1,");
	  if(a[i+j]=='x')
	    printf("x as 0,");
	  if(a[i+2*j]=='x')
	    printf("x as 1,");
          printf(")");
	}
	printf("  :Indexes are,%d,%d,%d\n",i,i+j,i+2*j);	//printing the index
      }

    }
}

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);
}