poj 1316 Self Numbers

朱雀 2022-05-29 04:19 141阅读 0赞

Self Numbers

<table> 
 <tbody> 
  <tr> 
   <td><strong>Time Limit:</strong>&nbsp;1000MS</td> 
   <td>&nbsp;</td> 
   <td><strong>Memory Limit:</strong>&nbsp;10000K</td> 
  </tr> 
  <tr> 
   <td><strong>Total Submissions:</strong>&nbsp;24659</td> 
   <td>&nbsp;</td> 
   <td><strong>Accepted:</strong>&nbsp;13803</td> 
  </tr> 
 </tbody> 
</table>

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

题目解析：

这个题目就是让求由自身加上其各位和得不到的数，直接暴力，1A

代码：

#include <iostream>
    using namespace std;
    const int maxn = 1e6;
    int a[maxn];
    int solve(int n)
    {
        int ans = 0;
        while(n)
        {
            ans+=(n%10);
            n/=10;
        }
        return ans;
    }
    
    int main()
    {
        for(int i=1;i<=10000;i++)
            a[i+solve(i)]=1;
        for(int i=1;i<=10000;i++)
            if(a[i]==0)
            cout<<i<<endl;
        return 0;
    }