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
|
|
|
   
   
   
   

 

//哈希水题

#include <stdio.h>
#include <string.h>

int hash[1000005];

int fun(int k)
{
    int sum = 0;
    while(k)
    {
        int r = k%10;
        sum+=r;
        k/=10;
    }
    return sum;
}
int main()
{
    int i;
    memset(hash,1,sizeof(hash));
    for(i = 1;i<=1000000;i++)
    {
        int sum = i;
        int k;
        sum += fun(i);
        hash[sum] = 0;
    }
    for(i = 1;i<=1000000;i++)
    {
        if(hash[i])
        printf("%d\n",i);
    }

    return 0;
}