ZOJ Problem Set–1180 Self Numbers

Time Limit: 10 Seconds      Memory Limit: 32768 KB

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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
|
|
|

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Source: Mid-Central USA 1998

筛选法：

#include<iostream>
#include<bitset>
using namespace std;
int d(int n)
{
  int number = n;
  while(n)
  {
    number += n%10;
    n /= 10;
  }
  return number;
}
bitset<1000001> bs;
int main()
{
  bs.set();
  int number = 0;
  for(int i = 1; i < bs.size(); i++)//删选不符合条件的数字
  {
    number = d(i);
    if(number < bs.size())
    {
      if(bs.test(number))
        bs.reset(number);
    }
    else
      break;
  }
  for(int i = 1; i < bs.size(); i++)
  {
    if(bs.test(i))
      cout<<i<<endl;
  }
  return 0;
}