POJ1316 Self Numbers

简单水题，不用打表，算出1~10000的self number，运用数组下标即可。

Self Numbers

Time Limit: 1000MS		Memory Limit: 10000K
Total Submissions: 21721		Accepted: 12231

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

Source

Mid-Central USA 1998

 

//oimonster
#include<cstdio>
#include<cstdlib>
#include<iostream>
using namespace std;
int a[20001];
int count(int i){
    int p=i;
    int s=i;
    while(p>0){
        s=s+p%10;
        p/=10;
    }
    return s;
}
int main(){
    int i,j,n;
    for(i=1;i<=20000;i++)a[i]=0;
    for(i=1;i<=10000;i++){
        a[count(i)]=1;
    }
    for(i=1;i<=10000;i++){
        if(a[i]==0)printf("%d\n",i);
    }
    return 0;
}