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.

Code

 1 #include <iostream>
 2 
 3 using namespace std;
 4 
 5 int main()
 6 { 
 7     bool b[10000]={false};
 8     int i=0,sum,d;
 9     for(i=0;i<10000;++i)
10     {
11         sum=d=i+1;
12         for(;d>0;d/=10)
13             sum+=d%10;
14         if(sum-1<10000)
15             b[sum-1]=true;
16     }
17     for(i=0;i<10000;++i)
18         if(!b[i])
19             cout<<i+1<<endl;
20     
21     return 0;
22 }

注：建立相应数组标记状态的方法值得学习。