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.
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\n",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