hdu 1333 Smith Numbers

Problem 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 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

49377740

 

Sample Output

4937775 

 

练练素数筛选

#include <iostream>#include <stdio.h>#include <string.h>#include <math.h>using namespace std;const int MAX=10001;bool prime[MAX];int getsum(int x){    int sum=0;    while (x)    {        sum+=x%10;        x/=10;    }    return sum;}int main(){    prime[2]=true;    int n;    for (int i=3; i<MAX; i++)    {        if(i%2==0)        {            prime[i]=false;        }        else        {            prime[i]=true;        }    }    for (int i=3; i<sqrt(MAX); i++)    {        if(prime[i])        {            for (int j=2; j*i<MAX; j++)            {                prime[j*i]=false;            }        }    }    while (scanf("%d",&n) && n)    {        int m;        while (n++)        {            int sum=0;            m=n;            for (int i=2; i<=sqrt(m); i++)            {                if(prime[i])                {                    if(m%i==0)                    {                        m/=i;                        int ans=getsum(i);                        sum+=ans;                        while(m%i==0)                        {                            m/=i;                            sum+=ans;                        }                    }                }            }            if(m==n)            {                continue;            }            if(m>1)            {                sum+=getsum(m);            }            if(getsum(n)==sum)            {                printf("%d\n",n);                break;            }        }    }    return 0;}