POJ 1316 Self Numbers(水题)

Self Numbers
Time Limit: 1000MS         Memory Limit: 10000K
Total Submissions: 20966         Accepted: 11775
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

题意：一个数n+自身每一个数和，组成一个新数。输出<=1000000不能够被这样组成的数。

分析：

暴力

#include <stdio.h>
#include <string.h>
#include <iostream>
#include <algorithm>
#include <vector>
#include <queue>
#include <set>
#include <map>
#include <string>
#include <math.h>
#include <stdlib.h>
#include <time.h>
using namespace std;

typedef long long ll;
const int N=10001000;
int vis[N];
ll f(ll n)
{
    ll sum=n;
    while(n)
    {
        sum+=n%10;
        n/=10;
    }
    return sum;
}
int main(){
    int t;
    for(int i=1;i<=1000000;i++)
    {
        vis[f(i)]=1;    
    }
    for(int i=1;i<=1000000;i++)
    {
        if(vis[i]==0)
            printf("%d\n",i);   
        //if(i>100) break;
    }
    return 0;
}