Self Numbers

Time Limit: 1000MS Memory Limit: 10000K
Total Submissions: 18560 Accepted: 10426

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.

Input

No input for this problem.

Output

Write a program to output all positive self-numbers less than 10000 in increasing order, one per line.

Sample Input


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
解:
这道题需要筛选出这些自数,遍历所有的10000以内的数,然后计算出每个数对应的生成数,给数组中对应的index标记上其为生成数的标识,最后通过这些标识筛选出自数。
#include <stdio.h>

int selfNum(int num)
{
    int check = num, sum = num;
    while (check != 0) {
        sum += check % 10;
        check /= 10;
    }
    return sum;
}
 
int main()
{
    char list[9998];
    int i;
    for (i = 1; i < 10000; i++) {
        int index = selfNum(i) - 1;
        if (index < 10000) {
            list[selfNum(i) - 1] = '1';
        }
    }
    for (i = 1; i < 10000; i++) {
        if (list[i - 1] != '1') {
			printf("%d\n", i);
        }
    }
    return 0;
}
Result Memory Time Language Code Length
Accepted 168K 0MS C++ 523B

 

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