HDU 1128 self numbers

Self Numbers

Time Limit: 20000/10000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 4451    Accepted Submission(s): 1943

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
|
|
|
   
   
   
   

 

一道暴力水题，直接一个循环处理出来就好~

#include <cstdio>
#include <stdlib.h>
#include <string.h>
#include <algorithm>
using namespace std;
bool a[1000005];

void solve( )
{
     for( int i=1 ; i<=1000000 ; i++ )
     {
         int t1 , t2;
         t1 = t2 = i;
         while( t1 )
         {
                t2 += t1 % 10;
                t1 /= 10;       
         }
         a[t2] = true;
     }
     for( int i=1 ; i<=1000000 ; i++ )
          if( !a[i] )
     printf("%d\n",i);
}

int main( )
{
    memset( a , false , sizeof(a) ); 
    solve( );
    return 0;
}