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.
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
Recommend
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 ",i); 30 } 31 return 0; 32 }