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.
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
Source
题目类型：模拟
算法分析：按照题目直接模拟即可
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1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36 37 38 39 40 41 42 43 44 45 |
#include <cstdio> #define N 10000 using namespace std; unsigned g[N]; unsigned sum_of_digits (unsigned n) { if (n < 10) return n; else return (n % 10) + sum_of_digits (n / 10); } void generate_sequence (unsigned n) { while (n < N) { unsigned next = n + sum_of_digits (n); if (next >= N || g[next] != next) return ; g[next] = n; n = next; } } int main() { unsigned n; for (n = 1; n < N; ++n) g[n] = n; for (n = 1; n < N; ++n) generate_sequence (n); for (n = 1; n < N; ++n) if (g[n] == n) printf ("%u\n", n); return 0; } |
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