1027: Smith Numbers

1027: Smith Numbers
Status
In/Out
TIME Limit
MEMORY Limit
Submit Times
Solved Users
JUDGE TYPE
stdin/stdout
30s
8192K
2332
568
Standard
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:
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 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

#include<stdio.h>
int b_s(int n)
{
 int l=0;
 while(n)
 {
  l+=n%10;
  n/=10;
 }
 return l;
}
int main()
{
 int n,k,sum2,b;
 while(scanf("%d",&n)==1)
 {
  while(n++)
  {
    b=n;k=2;sum2=0;
          while(k*k<=b)
    {
     if(b%k==0)
     {
      sum2+=b_s(k);
      b/=k;
     }
     while(b%k==0)
     {
      sum2+=b_s(k);
      b/=k;
     }
     k++;
    }
    if(b==n) continue;
    if(b>1&&b<n)
    {
     sum2+=b_s(b);
    } 
    if(b_s(n)==sum2)
    {
     printf("%d/n",n);
     break;
     
    }
  }
 }
  
 return 0;
}