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 descriptionin 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 (n )), 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 output135792031425364 | <-- a lot more numbers | 9903991499259927993899499960997199829993 |

 

Source mid-Central USA 1998 Nima, It's too simple, and the water goes through ...... code:

1 # include <cstdio> 2 # include <cstring> 3 # define maxn 1000001 4/* calculate the sum of single digits */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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