浙大在线评测 1180 Self Numbers

Problem:

    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
|
|
|


Solution:

#include <iostream>
using namespace std;

// 声明：本代码仅供学习之用，请不要作为个人的成绩提交。
// http://blog.csdn.net/mskia
// email: [email protected]

const long limit = 1000000;

int main( void ) {
    bool boo[ limit ];
    for ( long i = 0; i < limit; ++i ) {
        boo[ i ] = false;
    }
   
    for ( long i = 1; i <= limit; ++i ) {
        long t = i , r = i;
        while ( t > 0 ) {
            r += t % 10;
            t /= 10;
        }
       
        if ( r <= limit ) {
            boo[ r - 1 ] = true;
        }
    }
   
    for ( long i = 0; i < limit; ++i ) {
        if ( !boo[ i ] ) {
            cout << i + 1 << endl;
        }
    }

    system( "pause" );
    return 0;
}

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