poj1316,

## Problem:

Self Numbers
 Time Limit: 1000MS Memory Limit: 10000K Total Submissions: 14666 Accepted: 8252
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

## Solution:

Basically, it’s the most straight forward simulation problem. Just set all the non-self numbers as true, and then display the false ones.

## Source Code:

//Tue Apr 13 15:17:05 CDT 2010
#include <vector>
#include <list>
#include <map>
#include <set>
#include <deque>
#include <queue>
#include <stack>
#include <bitset>
#include <algorithm>
#include <functional>
#include <numeric>
#include <utility>
#include <sstream>
#include <iostream>
#include <iomanip>
#include <cstdio>
#include <cmath>
#include <cstdlib>
#include <cctype>
#include <string>
#include <cstring>
#include <cstdio>
#include <cmath>
#include <cstdlib>
#include <ctime>

using namespace std;
map<int, bool> mp;
int main(int argc, const char* argv[])
{
freopen("input.in", "r", stdin);
freopen("output.out", "w", stdout);
for (int i = 1; i <= 10000; i++)
{
stringstream s;
string str;
s << i;
str = s.str();
int sum = i;
for (int j = 0; j < (int) str.size(); j++)
{
sum += str[j] - '0';
}
mp[sum] = true;
if (mp[i] == false)
cout << i << endl;
}
fclose(stdin);
fclose(stdout);
return 0;
}