HDU - 1128 Self Numbers

题目：

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. 

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
|
|
|

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
|
|
|

这个题目，主要就是筛法的思想

代码：

#include<iostream>
using namespace std;

int list[1000001];

void d(int i)
{
	int x1 = i % 10, x2 = i / 10 % 10, x3 = i / 100 % 10;
	int x4 = i / 1000 % 10, x5 = i / 10000 % 10, x6 = i / 100000;
	i += x1 + x2 + x3 + x4 + x5 + x6;
	if (i <= 1000000)list[i] = 0;
}

int main()
{
	for (int i = 1; i <= 1000000; i++)list[i] = 1;
	for (int i = 1; i < 1000000; i++)d(i);
	for (int i = 1; i <= 1000000; i++)if (list[i])cout << i << endl;
	return 0;
}