Self Numbers
| Time Limit: 1000MS | Memory Limit: 10000K | |
| Total Submissions: 20864 | Accepted: 11709 |
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.
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.
Sample Input
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
这题第一眼看上去就觉得应该要用什么特殊方法，要不然会超时，可是最后没想到根本不会超时，我在调试中提交了n次Runtime error，让我几乎快疯了，后来当我把定义的bool数组移到main函数外就ac了，这让我百思不得其解，难道main函数内部定义一个10000长度的bool数组会超过限制吗
1 #include <iostream> 2 #include <cstdio> 3 #include <cstring> 4 using namespace std; 5 const int MAXn = 10000; 6 int numOf_Digit(int number) 7 { 8 int sum = number; 9 while(number>=10) 10 { 11 sum += number%10; 12 number /=10; 13 } 14 sum+=number ; 15 return sum; 16 } 17 bool vis[MAXn]; 18 int main() 19 { 20 21 memset(vis,true,sizeof(vis)); 22 for(int i=1;i<10000;i++) 23 { 24 int t =numOf_Digit(i); 25 vis[t] = false; 26 } 27 for(int i =1 ;i<10000;i++) 28 if(vis[i]==true) 29 printf("%d ",i); 30 return 0; 31 }
当然直接把#define 去掉，直接就是vis[10000],并且取消掉那个运算函数，用一句int t = i + i/1000+(i/100)%10+(i/10)%10+i%10; 就可以0MS了，但个人风格，喜欢把功能变成函数