ACM Self Number

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
题解：
　　　基础题。题目不难，只要能够理解题意的话就能够顺利的做出来了.
　　　　
代码：

 /*
     Name: Self Number
     Copyright:
     Author:
     Date: 11/08/17 04:23
     Description:
 */
 #include<stdio.h>
 #include<iostream>
 #include<cstring>
 #include<algorithm>
 using namespace std;
 const int NUM = ;
 int record[NUM];
 int judge(int i)
 {
     int temp = i;
     while(i)
     {
         temp += i%;
         i /= ;
     }
     return temp;   /*返回非self number*/
 }

 int main()
 {
     memset(record,,sizeof(record));  /*数组初始化，是一个好习惯*/
     for(int i = ; i < NUM;i++)
     {
         int  temp  = judge(i);
         if(temp < NUM )
             record[temp] = ;    /*将不是self number的整数标记为1*/
     }
     bool flag = false;
     for(int i = ; i < NUM;i++)
     {

         if(record[i] ==  )/*注意输出格式 最后一个不空行*/
         {
             if(flag)
                 cout<<endl;
             cout<<i;
             flag = true;
         }

     }
 return ;
 }