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

135792031425364 | |       <-- a lot more numbers |9903991499259927993899499960997199829993

来源

Mid-Central USA 1998

3.思路：

4.代码：

1 //2010-04-28 2 //v0.1 create by wuzhihui 3 #include
     
       4 using namespace std; 5 #define max 10000 6 int a[max+2]={
   0}; 7  8 int main() 9 {10     int b,c;11     int i;12     //memset(a,1,sizeof(a));13     for(i=1;i<=max;i++)14     {15         b=c=i;16         do17         {18               b+=(c%10);19               c=c/10;20         }while(c!=0);21         if(b<=max)  a[b]=1;22     }23     for(i=1;i<=max;i++)24     {25        if(a[i]==0) cout<
      <