HDU 1128 Self Numbers

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分类专栏： hdu ACM 水题128题文章标签： numbers generator integer output less ini

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本文探讨了自数字的概念、生成方法及识别算法，通过构建序列并标记非自数字，最终输出所有小于等于1000000的正自数字。

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Self Numbers

Time Limit: 20000/10000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 4596 Accepted Submission(s): 2020

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

Recommend

Eddy

分析：因为若这个不是selfnumber那他肯定有generator，所以我们的想法就是把每个数都当做generator，求他们的后一个数，则求出来

的这个数做个标记，表示一定不是selfnumber，到最后就可以筛选出来那些数字是selfnumber

代码：

#include<stdio.h>
#define N 1000001
bool str[N];

void func(int n)
{
    int sum=n;
    while(n)
    {
        sum+=n%10;
        n/=10;
    }
    str[sum]=1;
}

int main()
{
    int i;
    for(i=1;i<N;i++)
    func(i);
    for(i=1;i<N;i++)
    {
        if(!str[i])
            printf("%d\n",i);
    }
    return 0;
}