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+2]={0};

int main()
{
    int b,c;
    int i;
    //memset(a,1,sizeof(a));
    for(i=1;i<=max;i++)
    {
        b=c=i;
        do
        {
              b+=(c%10);
              c=c/10;
        }while(c!=0);
        if(b<=max)  a[b]=1;
    }
    for(i=1;i<=max;i++)
    {
       if(a[i]==0) cout<<i<<endl;
    }
    //system("pause");
    return 1;
}