POJ1316 Self Numbers

Self Numbers

Time Limit: 1000MS Memory Limit: 10000KTotal Submissions: 22211 Accepted: 12496

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

135792031425364 | |       <-- a lot more numbers |9903991499259927993899499960997199829993

Source

Mid-Central USA 1998

#include<iostream>#include<cstdio>using namespace std;int a[10001]={0};int main(){    int i,j,k,w;    for(i=1;i<=10000;i++)    {        j=i;        w=i;        while(w>0)//算出各个位数之和，一旦有这样的数便不是自数，先剔除出去        {            k=w%10;            j+=k;            w/=10;        }        a[j]=1;    }    for(i=1;i<=10000;i++)    {        if(a[i]!=1)            printf("%d\n",i);    }    return 0;}