HDU1128:Self Numbers

Problem 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. 


Write a program to output all positive self-numbers less than or equal 1000000 in increasing order, one per line. 

 

Sample Output

135792031425364|| <-- a lot more numbers|9903991499259927993899499960997199829993||
|
tip:一道水题....
#include<iostream>#include<cstring>using namespace std;int _hash[1000005];int fun(int x){int sum=0;while(x/10){sum+=x%10;x/=10;}sum+=x;return sum; } int main(){memset(_hash,0,sizeof(_hash));for(int i=1;i<=1000000;i++){int sum=i+fun(i);_hash[sum]=1; }for(int i=1;i<=1000000;i++)if(!_hash[i])cout<<i<<endl; return 0; }