POJ1316Self Numbers 注意數組範圍

Self Numbers

Time Limit: 1000MS		Memory Limit: 10000K
Total Submissions: 23444		Accepted: 13170

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

求所有的自己數，就是不能通過別的數字到達的數

#include <iostream>
#include <stdio.h>
#include <string>
#include <cstring>
#include <algorithm>
#include <cmath>

using namespace std;
int vis[20009];
int main()
{
    memset(vis,0,sizeof vis);
    for(int i=1;i<=10000;i++)
    {
            int tmp=i;
            int t,f;
            while(tmp<=10000)
            {
                f=tmp;
                t=f;
                while(f)
                {
                    t+=f%10;
                    f/=10;
                }
                vis[t]=1; //沒判斷t的範圍，此時可能超過10000，數組要開大，或者判斷下大小
                tmp=t;
            }
    }

    for(int i=1;i<=10000;i++)
    {
        if(vis[i]==0)
        printf("%d\n",i);
    }


    return 0;
}