Zoj problem set-1180 self numbers

Time Limit: 10 seconds memory limit: 32768 KB

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 (n )), 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

Signature method:

# Include <iostream>
 
 
# Include <bitset>
 
 
Using NamespaceSTD;
 
 
 
 
 
IntD (IntN)
 
 
{
 
 
IntNumber = N;
 
 
While(N)
 
 
{
 
Number + = n % 10;
 
 
N/= 10;
 
 
}
 
 
ReturnNumber;
 
 
}
 
Bitset <1000001> BS;
 
 
IntMain ()
 
 
{
 
 
 
 
BS. Set ();
 
IntNumber = 0;
 
 
For(IntI = 1; I <BS. Size (); I ++)// Delete and select numbers that do not meet the conditions
 
 
{
 
 
Number = d (I );
 
If(Number <BS. Size ())
 
 
{
 
 
If(BS. Test (number ))
 
 
BS. Reset (number );
 
 
}
 
Else
 
 
Break;
 
 
}
 
 
For(IntI = 1; I <BS. Size (); I ++)
 
 
{
 
If(BS. Test (I ))
 
 
Cout <I <Endl;
 
 
}
 
 
Return0;
 
 
}