HDU 1128Self Numbers(筛选)

Self Numbers

Time Limit: 20000/10000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 5462    Accepted Submission(s): 2414

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

 

 

              题目大意： 输出1~1000000中不是自循环的数字，所谓自循环就是可以由由别的数通过每个数位相加在加上原来的数字得到。

        解题思路：由于该题是求一个数是不是“自身数”，而定义一个数是不是“自身数”的根据就是是否有祖先，又因为一个数的祖先一定比这个数要小，所以这就和筛选法很像了。直接筛选，开始因为数组开小了，老是爆栈。这个题目没有输入哈哈。

      题目地址：Self Numbers

AC代码：

#include<iostream>
#include<cstring>
#include<cstdio>
using namespace std;
int maxn=1000000;
bool a[2000005];

int main()
{
    int i,p;
    for(i=1;i<=maxn;i++) //难怪会爆栈
    {
        p=i;
        int tmp = i;
        while(tmp)
        {
           p+=tmp%10;
           tmp/=10;
        }
        a[p]=true;
    }

    for(i=1;i<=maxn;i++)
        if(!a[i])
        {
           printf("%d\n",i);
        }
    return 0;
}
//78MS 1204K