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.
Code
1 #include <iostream>
2
3 using namespace std;
4
5 int main()
6 {
7 bool b[
10000]={
false};
8 int i=
0,sum,d;
9 for(i=
0;i<
10000;++
i)
10 {
11 sum=d=i+
1;
12 for(;d>
0;d/=
10)
13 sum+=d%
10;
14 if(sum-
1<
10000)
15 b[sum-
1]=
true;
16 }
17 for(i=
0;i<
10000;++
i)
18 if(!
b[i])
19 cout<<i+
1<<
endl;
20
21 return 0;
22 }
注：建立相应数组标记状态的方法值得学习。
转载于:https://www.cnblogs.com/QuentinYo/archive/2013/04/04/2999459.html
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