Smith Numbers（分解质因数）

Smith Numbers

Time Limit: 1000MS		Memory Limit: 10000K
Total Submissions: 14173		Accepted: 4838

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

Source

Mid-Central European Regional Contest 2000

AC代码:

 1 #include<iostream>
 2
 3 using namespace std;
 4
 5 int CalDigitsSum(int num)
 6 {
 7     int sum = 0;
 8     while(num)
 9     {
10         sum += num % 10;
11         num /= 10;
12     }
13     return sum;
14 }
15
16 int PrimaryCal(int num)
17 {
18     int total = 0;
19     int tempNum = num;
20     for(int i = 2; i * i <= num; i++)
21     {
22         int temp;
23         if(num % i == 0)
24             temp = CalDigitsSum(i);
25         while(num % i == 0)
26         {
27             total += temp;
28             num /= i;
29         }
30     }
31     if(tempNum == num)
32         return -1;
33     if(num != 1)
34         total += CalDigitsSum(num);
35     return total;
36 }
37
38 int main()
39 {
40     int n;
41     while(1)
42     {
43         cin >> n;
44         if(n == 0)
45             break;
46         for(int i = n + 1; ; i++)
47         {
48             if(CalDigitsSum(i) == PrimaryCal(i))
49             {
50                 cout << i << endl;
51                 break;
52             }
53         }
54     }
55     return 0;
56 }