poj 1142 Smith Numbers

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

#include<iostream>
#include<cstdio>
#include<cstring>
#include<cmath>
#include<queue>
using namespace std;

int distsum(int n)
{
    int ans=0;
    while(n)
    {
        ans+=n%10;
        n=n/10;
    }
    return ans;
}

bool isprime(int n)
{
    if(n==1) return false;
    if(n==2) return true;
    for(int i=2;i<=(int)sqrt(n+0.5)+1;i++)
    {
        if(n%i==0)
        return false;
    }
    return true;
}

int prime_factor(int n)
{
    int i=2;
    queue <int> q;
    while(n!=1||n!=0)
    {
        if(n%i==0&&isprime(i))
        {
            q.push(i);
            n/=i;
            if(isprime(n))
            {
                q.push(n);break;
            }
        }
        else i++;
    }
   
    while(!q.empty())
    {
        int k=q.front();
        q.pop();
        cout<<k<<endl;
    }
    return 0;
}

int main()
{
    int n;
    while(cin>>n)
    {
        if(n==0) break;
        for(int i=n+1;;i++)
        {
            if(isprime(i)) continue;
            if(prime_factor_sum(i)==distsum(i))
            {
                cout<<i<<endl;break;
            }
        }
    }
    return 0;
}