poj1316 Self Numbers

Self Numbers

Time Limit: 1000MS		Memory Limit: 10000K
Total Submissions: 16385		Accepted: 9189

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 <iostream>
using namespace std;
bool s[10001]={0};
int d(int k)
{
int n=k;
while(n)
{
k+=n%10;
n/=10;
}
return k;
}
int main()
{
int i,j;
for(i=1;i<=10000;++i)
{
j=d(i);
if(j>10000)
continue;
s[j]=1;
}
for(i=1;i<=10000;++i)
if(!s[i])
cout<<i<<endl;
return 0;
}