Self Numbers

Time Limit: 20000/10000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 4487    Accepted Submission(s): 1961


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 | | |
 
 
 1 #include<stdio.h>
 2 int flag[1000001];
 3 
 4 int fun(int i)
 5 {
 6     int t=i;
 7     while(t)
 8     {
 9         i+=t%10;
10         t/=10;
11     }
12     return i;
13 }
14 
15 int main()
16 {
17     int i;
18     //freopen("1.text","w",stdout); 
19     for(i=1;i<1000001;++i)
20     {
21         flag[fun(i)]=1;
22     }
23     for(i=1;i<1000001;++i)
24         if(flag[i]==0)
25             printf("%d\n",i);
26         return 0;
27 }

 

posted on 2012-08-20 09:26  可笑痴狂  阅读(1542)  评论(0编辑  收藏  举报