hdu 1128 Self Numbers

Self Numbers

Time Limit: 20000/10000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 6960    Accepted Submission(s):
3047

Problem 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.

Write a
program to output all positive self-numbers less than or equal 1000000 in
increasing order, one per line.

 

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

Mid-Central
USA 1998

 

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这道题没有输入，看题目的意思写这个数组，直接筛选暴力打表就好。

 

题意：输出这一串数字，1000000以内所有的每个数，加上自身每一位数字，可以生产另一个数，但有些数，不能通过这样产生，请输出这些数。

 

附上代码：

 

 #include <iostream>
 #include <cstdio>
 #include <cstring>
 using namespace std;
 int visit[];
 int xx(int n)
 {
     int sum=;
     while(n!=)
     {
         sum+=n%;
         n/=;
     }
     return sum;
 }
 int main()
 {
     int i,j,sum;
     memset(visit,,sizeof(visit));
     for(i=; i<=; i++)   //直接暴力打表
     {
         sum=i;
         sum+=xx(i);
         visit[sum]=;         //不需要出现的数字标记为0
     }
     for(i = ; i<=; i++)
     {
         if(visit[i])
             printf("%d\n",i);
     }
     return ;
 }