POJ 1316 Self Numbers

Self Numbers

Time Limit: 1000MS		Memory Limit: 10000K
Total Submissions: 18220		Accepted: 10262

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

Source
Mid-Central USA 1998
解题思路：水题而已，从反面考虑，枚举1到10000的所有数可以组成哪些SelfNumber，再把这些数剔除即是答案。注意枚举的时候数组开大一点，例如9999这个数可以变成9+9+9+9+9999=10035，因此数组开10100即可。

#include<iostream>
using namespace std;
int main()
{
	int i,j,k,obj,temp;
	bool selfnumber[10100];
	int num[6];
	memset(selfnumber,true,sizeof(selfnumber));
	for(i=1;i<=10000;i++)
	{
		memset(num,0,sizeof(num));
		obj=0;
		num[5]=i;
		temp=i;
		for(j=0;j<5;j++)
			{num[j]=temp%10;
		    temp/=10;}
		for(k=0;k<6;k++)
			obj+=num[k];
		selfnumber[obj]=false;
	}
	for(i=1;i<10001;i++)
		if(selfnumber[i])
			cout<<i<<endl;
}