Self Numbers
Time Limit: 20000/10000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 6960 Accepted Submission(s):
3047
Problem 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
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
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这道题没有输入，看题目的意思写这个数组，直接筛选暴力打表就好。
题意：输出这一串数字，1000000以内所有的每个数，加上自身每一位数字，可以生产另一个数，但有些数，不能通过这样产生，请输出这些数。
附上代码：
1 #include <iostream>
2 #include <cstdio>
3 #include <cstring>
4 using namespace std;
5 int visit[1000001];
6 int xx(int n)
7 {
8 int sum=0;
9 while(n!=0)
10 {
11 sum+=n%10;
12 n/=10;
13 }
14 return sum;
15 }
16 int main()
17 {
18 int i,j,sum;
19 memset(visit,1,sizeof(visit));
20 for(i=1; i<=1000000; i++) //直接暴力打表
21 {
22 sum=i;
23 sum+=xx(i);
24 visit[sum]=0; //不需要出现的数字标记为0
25 }
26 for(i = 1; i<=1000000; i++)
27 {
28 if(visit[i])
29 printf("%d\n",i);
30 }
31 return 0;
32 }