POJ 1316 Self Numbers

题目链接：

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

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. 

Input

No input for this problem.

Output

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

Sample Input

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
Hint
题意：

就是让你求1到10000以内所有的自私数，什么叫自私数？如果一个数不能分解为另一个数和那个数各位数字之和，它就是一个"自私数"，举个例子，57可以是51+5+1来得到，那么57就不是自私数。那么100以内的自私数是：1，3，5，7，9，20，31，42，53，64，75，86和97。

题解：

直接把1到10000的非自私数打表存起来就行了。

代码：

#include <cmath>
#include <cstdio>
#include <cstring>
#include <algorithm>
using namespace std;
#define met(a,b) memset(a,b,sizeof(a))
const int maxn = 1e4;
int a[maxn];
int main()
{
    met(a,0);
    for(int i=1;i<maxn;i++)
    {
        int num=i;
        int sum=num;
        while(num)
        {
            sum+=num%10;
            num/=10;
        }
        if(sum<maxn)
            a[sum]=1;
    }
    for(int i=1;i<maxn;i++)
    {
        if(a[i]==0)
            printf("%d\n",i);
    }
}