HDU1128:Self Numbers

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. 

Write a program to output all positive self-numbers less than or equal 1000000 in increasing order, one per line. 

 

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
|
|
|

tip:一道水题....

#include<iostream>
#include<cstring>
using namespace std;

int _hash[1000005];
int fun(int x)
{
int sum=0;
while(x/10)
{
sum+=x%10;
x/=10;
}
sum+=x;
return sum;
}
int main()
{
memset(_hash,0,sizeof(_hash));

for(int i=1;i<=1000000;i++)
{
int sum=i+fun(i);
_hash[sum]=1;

}
for(int i=1;i<=1000000;i++)
if(!_hash[i])cout<<i<<endl;

return 0;
}