Smith Numbers

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

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给一个数，找到比这个数大的数x，x需要满足各位数和与其分解质因子之后的位数和相同
设f(x)为x的位数和

则 f(x)=f(x的因子) f ( x ) = f ( x 的 因 子 )

eg:

4937775=3∗5∗5∗65837 4937775 = 3 ∗ 5 ∗ 5 ∗ 65837

4+9+3+7+7+7+5=42 4 + 9 + 3 + 7 + 7 + 7 + 5 = 42

3+5+5+6+5+8+3+7=42 3 + 5 + 5 + 6 + 5 + 8 + 3 + 7 = 42

---

code

//#include
#include
#include
#include
#include
using  namespace std;
typedef long long ll;
const int MAXN=1e5+5;
int prime[MAXN+1];
int factor[1000][2];
int isprime(int n)
{
    for(int i=2;i*i<=n;i++)
        if(n%i==0)  return 0;
    return n>1;
}
void getprime()
{
    memset(prime,0,sizeof(prime));
    for(int i=2;i<=MAXN;i++)
    {
        if(!prime[i])   prime[++prime[0]]=i;
        for(int j=2;j*i<=MAXN;j++)
            prime[i*j]=1;
    }
}
int getfactor(int n)
{
    int cnt=0;
    memset(factor,0,sizeof(factor));
    for(int i=1;prime[i]*prime[i]<=n;i++)
    {
        factor[cnt][1]=0;//第二个元素记录幂 
        if(n%prime[i]==0)
        {
            factor[cnt][0]=prime[i];
            while(n%prime[i]==0)
            {
                n/=prime[i];
                factor[cnt][1]++;
            }
            cnt++;
        }
    }
    if(n!=1)
    {
        factor[cnt][0]=n;
        factor[cnt++][1]=1;
    }
    return cnt;
}
int f(int n)
{
    int ans=0;
    while(n)
    {
        ans+=n%10;
        n/=10;
    }
    return ans;
}
int cnt(int n)
{
    if(isprime(n))
        return f(n);
    for(int i=(int)sqrt(n+0.5);i>1;i--)
        if(n%i==0)  return cnt(i)+cnt(n/i);
 } 
int main()
{
    int n;
    getprime();
    while(~scanf("%d",&n)&&n)
    {
        int m=n;
        while(m++)
        {
            if(isprime(m)) continue;//这个数还不能是素数 
            int ans1=f(m);
            int ans2=cnt(m);
            if(ans1==ans2)
            {
                printf("%d\n",m);
                break;
            }
        }

    }
    return 0;
}