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
1 //oimonster
2 #include<cstdio>
3 #include<cstdlib>
4 #include<iostream>
5 using namespace std;
6 int a[20001];
7 int count(int i){
8 int p=i;
9 int s=i;
10 while(p>0){
11 s=s+p%10;
12 p/=10;
13 }
14 return s;
15 }
16 int main(){
17 int i,j,n;
18 for(i=1;i<=20000;i++)a[i]=0;
19 for(i=1;i<=10000;i++){
20 a[count(i)]=1;
21 }
22 for(i=1;i<=10000;i++){
23 if(a[i]==0)printf("%d\n",i);
24 }
25 return 0;
26 }