Self Numbers

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
 |
 |
 |

---

#include<stdio.h>
#include<stdlib.h>
#include<string.h>
char x[1000001];
int main() {
    memset(x, 0, sizeof(x));
    int i, j, tmp;
    for(i = 1; i <= 999999; i++) {
        j = i, tmp = 0;
        while(j) {
            tmp += j%10;
            j /= 10;
        }
        if(tmp+i <= 1000000)
            x[tmp+i] = 1;
    }
    for(i = 1; i <= 1000000; i++)
        if(!x[i])
            printf("%d\n", i);
    return 0;
}

---

以下作法會更快

#include<stdio.h>
#include<string.h>
char x[1000001], y[1000001];
int main() {
    memset(x, 0, sizeof(x));
    memset(y, 0, sizeof(y));
    int i, j, ti;
    for(i = 0; i <= 999999; i++) {
        if(i < 100000) {
            for(j = 0, ti = 10*i; j < 10; j++)
                y[ti+j] = y[i]+j;
        }
        if(i+y[i] <= 1000000)
            x[i+y[i]] = 1;
    }
    for(i = 1; i <= 1000000; i++)
        if(!x[i])
            printf("%d\n", i);
    return 0;
}