hdoj-1128-Self Numbers

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|||                

妥妥的水题啊，题意就是求一堆的Self Numbers，它的定义是一个函数d(33)=33+3+3=39,d(75)=7+5+75=87，然后让你求所有不超过1000000的这个Self Numbers。

#include<cstdio>#include<cstring>#include<algorithm>#include<iostream>using namespace std;int a[1000005];void init(){    for(int i=1;i<=1000000;i++)    {        int k=i;        int sum=i;        int n;        while(k)        {            n=k%10;            sum+=n;            k=k/10;        }        a[sum]=0;    }}int main(){    memset(a,1,sizeof(a));    init();    for(int i=1;i<=1000000;i++)    {        if(a[i])        printf("%d\n",i);        //system("pause");    }}