# Self Numbers

 Time Limit: 1000MS Memory Limit: 10000K Total Submissions: 18560 Accepted: 10426

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

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 selfNum(int num)
{
int check = num, sum = num;
while (check != 0) {
sum += check % 10;
check /= 10;
}
return sum;
}

int main()
{
char list[9998];
int i;
for (i = 1; i < 10000; i++) {
int index = selfNum(i) - 1;
if (index < 10000) {
list[selfNum(i) - 1] = '1';
}
}
for (i = 1; i < 10000; i++) {
if (list[i - 1] != '1') {
printf("%d\n", i);
}
}
return 0;
}```
 Result Memory Time Language Code Length Accepted 168K 0MS C++ 523B

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