【思维/模拟】Self Numbers

传送门:POJ1316
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.

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
|
|
|

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
|
|
|

翻译:

一个数加上其各位上的和,组成一个新数。
输出1-1000000中不能够被这样组成的数。
分析:桶思想打表
代码:
#include<cstdio>
#include<cstring>
using namespace std;
bool num[1000020];
int main()
{
  for (int i = 1; i <= 1000000; i++)
  {
    int sum = i / 1000000  +i / 100000 % 10 +i / 10000 % 10 +i / 1000%10 + i / 100 % 10 + i / 10 % 10 + i % 10 + i;
    num[sum] = true;
    if (!num[i])
      printf("%d\n", i);
  }
}

 

 

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