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HDUoj-------(1128)Self Numbers

Self Numbers

Time Limit: 20000/10000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others) Total Submission(s): 6227    Accepted Submission(s): 2728

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

Mid-Central USA 1998

尼玛,太简单了,之间就水过去了.....

代码:

 1 #include<cstdio>
 2 #include<cstring>
 3 #define maxn 1000001
 4 /*求个位数之和*/
 5 int work(int n)
 6 {
 7     int sum=0;
 8     while(n>0){
 9       sum+=n%10;
10       n/=10;
11     }
12     return sum;
13 }
14 bool ans[maxn];
15 int main(){
16     int pos;
17     //freopen("test.out","w",stdout);
18     memset(ans,0,sizeof(ans));
19     for(int i=1;i<maxn;i++){
20         pos=i+work(i);
21         if(pos<=1000000&&!ans[pos]) ans[pos]=1;
22     }
23     for(int i=1;i<maxn;i++){
24         if(!ans[i])printf("%d\n",i);
25     }
26 return 0;
27 }

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