Problem description
In 1949 the Indian mathematician D.R. Kaprekar discovered a class of numbers called self-numbers. For any positive integer n, define d(n) to be n plus the sum of the digits of n. (The d stands for digitadition, a term coined by Kaprekar.) For example, d(75) = 75 + 7 + 5 = 87. Given any positive integer n as a starting point, you can construct the infinite increasing sequence of integers n, d(n), d(d(n)), d(d(d(n))), .... For example, if you start with 33, the next number is 33 + 3 + 3 = 39, the next is 39 + 3 + 9 = 51, the next is 51 + 5 + 1 = 57, and so you generate the sequence 33, 39, 51, 57, 69, 84, 96, 111, 114, 120, 123, 129, 141, ... The number n is called a generator of d(n). In the sequence above, 33 is a generator of 39, 39 is a generator of 51, 51 is a generator of 57, and so on. Some numbers have more than one generator: for example, 101 has two generators, 91 and 100. A number with no generators is a self-number. There are thirteen self-numbers less than 100: 1, 3, 5, 7, 9, 20, 31, 42, 53, 64, 75, 86, and 97.

 
Input
No input for this problem.

 
Output
Write a program to output all positive self-numbers less than 10000 in increasing order, one per line.

 
Sample Input
no input
 
Sample Output
1
3
5
7
9
20
31
42
53
64
 |
 |       <-- a lot more numbers
 |
9903
9914
9925
9927
9938
9949
9960
9971
9982
9993
 

#include <stdio.h>

int main()
{int test,a,b,c,d,m;

 printf("1\n3\n5\n7\n9\n");
 for(m = 20; m < 10000; m ++)
   { if( m > 10 && m <= 117)
 { for(a = 1; a < 10; a ++)
     for(b = 0; b < 10; b ++)
  { test = a * 10 + b + a + b;
    if( test == m )
   { test = -1;
     goto next;
   }
  }
 }
    if( m >= 100 && m <= 1026 )
 { for(a = 1; a < 10; a ++)
     for(b = 0; b < 10; b ++)
  for(c = 0; c < 10; c ++)
  { test = a * 100 + b * 10 + c + a + b + c ;
    if( test == m )
   { test = -1;
     goto next;
   }
  }
 }
   if( m >= 1000 )
 { for(a = 1; a < 10; a ++)
     for(b = 0; b < 10; b ++)
       for(c = 0; c < 10; c ++)
   for(d = 0; d < 10; d ++)
  { test = a * 1000 + b * 100 + c * 10 + d + a + b + c + d;
    if( test == m )
   { test = -1;
     goto next;
   }
  }
 }
   next:    if( test != -1 )
  printf("%d\n",m);
  }
 return 0;
}

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