POJ 1316 Self Numbers

Self Numbers

Time Limit: 1000MS Memory Limit: 10000KTotal Submissions: 21832 Accepted: 12291

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

要求无前驱的数，用类似于筛法求素数，我们只需要把每个数的后继删掉，那么剩下的肯定是无前驱的。（可以有反证法简单证明一下：如果剩下的数有前驱，那么这个数就是其前驱的后继，那么一定会被删掉，所以不会剩下，故矛盾，即剩下的数无前驱，也就是我们要求的）

#include <stdio.h>#include <string.h>#include <math.h>#include <algorithm>using namespace std;#define eps 1e-9#define max(a,b) (a)>(b)?(a):(b)int vis[11010];int main(){int i,j,k,n,res,t;memset(vis,0,sizeof(vis));vis[0]=vis[1]=0;for(i=1;i<=10000;i++){n=t=i;while(n){t+=(n%10);n/=10;}vis[t]=1;}for(i=1;i<=10000;i++)if(!vis[i])printf("%d\n",i);return 0;}