Self Numbers – POJ 1316 水题

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Self Numbers

Time Limit:

1000MS

Memory Limit:

10000K

Total Submissions:

20966

Accepted:

11775

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


题意:让一个数加上其各位上的和,组成一个新数。输出不能够被这样组成的数。


原谅我在回家等火车的时候闲来无事切切水题………………


AC代码如下:


#include<cstdio>
#include<cstring>
using namespace std;
int num[20010];
int main()
{ int i,j,k;
  for(i=1;i<=10000;i++)
  { num[i/1000+i/100%10+i/10%10+i%10+i]=1;
    if(!num[i])
     printf("%d\n",i);
  }
}



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