POJ1142Smith Numbers一道简单的数学题

分类：数论

作者： ACShiryu

时间：2011-8-3

原题： http://poj.org/problem?id=1142

Smith Numbers

Time Limit: 1000MS		Memory Limit: 10000K
Total Submissions: 8853		Accepted: 3105

Description

While skimming his phone directory in 1982, Albert Wilansky, a mathematician of Lehigh University,noticed that the telephone number of his brother-in-law H. Smith had the following peculiar property: The sum of the digits of that number was equal to the sum of the digits of the prime factors of that number. Got it? Smith's telephone number was 493-7775. This number can be written as the product of its prime factors in the following way: 
4937775= 3*5*5*65837
The sum of all digits of the telephone number is 4+9+3+7+7+7+5= 42,and the sum of the digits of its prime factors is equally 3+5+5+6+5+8+3+7=42. Wilansky was so amazed by his discovery that he named this kind of numbers after his brother-in-law: Smith numbers. 
As this observation is also true for every prime number, Wilansky decided later that a (simple and unsophisticated) prime number is not worth being a Smith number, so he excluded them from the definition. 
Wilansky published an article about Smith numbers in the Two Year College Mathematics Journal and was able to present a whole collection of different Smith numbers: For example, 9985 is a Smith number and so is 6036. However,Wilansky was not able to find a Smith number that was larger than the telephone number of his brother-in-law. It is your task to find Smith numbers that are larger than 4937775!

Input

The input file consists of a sequence of positive integers, one integer per line. Each integer will have at most 8 digits. The input is terminated by a line containing the number 0.

Output

For every number n > 0 in the input, you are to compute the smallest Smith number which is larger than n,and print it on a line by itself. You can assume that such a number exists.

Sample Input

4937774
0

Sample Output

4937775

题目大意就是要你找一个大于n的和数，并且满足他的各位的和与他所有的质因子的各位的和相等的最小的一个数，即此题的Smith Number

例如对于4937774，比他大的第一个数是4937775

因为4+9+3+7+7+7+5=42

又4937775=3*5*5*65837

而3+5+5+6+5+8+3+7=42

故4937775是题目要求的答案。

这是一道纯数学题，可以通过暴力直接得到答案，因为这样的数分布比较密，

不过在做这题时学到了很好的一个思想，分治法，详见代码；

提交1次就A了，刚开始时想复杂了，准备筛选做的，看了一下讨论，好像没必要，可直接暴力

参考代码：

 1 #include<iostream>
 2 #include<cstdlib>
 3 #include<cstdio>
 4 #include<cstring>
 5 #include<algorithm>
 6 #include<cmath>
 7 using namespace std;
 8 bool isprime (int k)
 9 {//判断是否是素数
10     int t = sqrt ( k + 0.5 );
11     for ( int i = 2 ; i <= t ; i ++ )
12         if ( k % i == 0 )
13             return false;
14     return true;
15 }
16 int sum(int k)
17 {//求出该数各位上的和
18     int i , s;
19     s = 0 ;
20     while ( k != 0 )
21     {
22         i = k % 10 ;
23         s += i ;
24         k = k / 10 ;
25     }
26     return s;
27 }
28 int cut (int k )
29 {//分治法思想，如果是素数，就返回sum，否则，就将该数分成两部分，再来求各部分的质因子的sum
30     if ( isprime(k) )
31         return sum (k);
32     for ( int i = (int) sqrt (k + 0.5) ; i >1 ; i -- )
33         if ( k % i == 0 ) 
34             return cut (i) + cut (k / i) ;
35 }
36 int main()
37 {
38     int n;
39     while ( cin >> n , n )
40     {
41         while ( n ++ )
42         {
43             if (!isprime(n)&&sum(n)==cut(n))
44                 break;
45         }
46         cout<<n<<endl;
47     }
48 }