Self Numbers


       Time Limit: 20000/10000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
                       Total Submission(s): 6156    Accepted Submission(s): 2688


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 | | |


 



Source


​​Mid-Central USA 1998​​​
题意：如题目所举的例子，就拿75来说，因为75+7+5=87，所以75是87的祖先，同理，33是39的祖先。
但是有的数祖先不一定只有一个数，比如100，91都是101的祖先。
因此，推导出自身数的定义：没有一个祖先的数为自身数，设计程序，打印1到100^3的自身数。
思路：自己首先想到筛法求素数的方法，利用数组下标标记。 

/***************
Author:jiabeimuwei
Sources:HDU1128
Times:203ms
***************/

#include<cstring>
#include<cstdlib>
#include<cstdio>
using namespace std;
const long long int n=1000000;
bool a[n]={0};
long long f(long long n)
{
    long long s=n;
    while(n)
    {
        long long int c=n%10;
        n/=10;
        s+=c;
    }
  return s;
}
int main()
{
    long long int i,j;
    for(i=1; i<=n; i++)
    {
        a[f(i)]=1;
    }
    for(i=1; i<=n; i++)
        if(!a[i])
            printf("%d\n",i);
        }