OpenJudge/Poj 1316 Self Numbers

1.链接地址：

http://poj.org/problem?id=1316

http://bailian.openjudge.cn/practice/1316

2.题目：

总时间限制:

1000ms

内存限制:

65536kB

描述

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.

输入

No input for this problem.

输出

Write a program to output all positive self-numbers less than 10000 in increasing order, one per line.

样例输入

样例输出

1
3
5
7
9
20
31
42
53
64
 |
 |       <-- a lot more numbers
 |
9903
9914
9925
9927
9938
9949
9960
9971
9982
9993

来源

Mid-Central USA 1998

3.思路：

4.代码：

 //2010-04-28
 //v0.1 create by wuzhihui
 #include<iostream>
 using namespace std;
 #define max 10000
 int a[max+]={};

 int main()
 {
     int b,c;
     int i;
     //memset(a,1,sizeof(a));
     for(i=;i<=max;i++)
     {
         b=c=i;
         do
         {
               b+=(c%);
               c=c/;
         }while(c!=);
         if(b<=max)  a[b]=;
     }
     for(i=;i<=max;i++)
     {
        if(a[i]==) cout<<i<<endl;
     }
     //system("pause");
     return ;
 }