hdu 1333 Smith Numbers（暴力思路）

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!

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.

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.

  
4937774
0

  
4937775 

import java.util.*;
public class Main {static int getsum(int x){int ans=0;while(x>0){ans=ans+x%10;x/=10;}return ans;}static final int maxn=(int)(1e4+10);static int[] fac=new int [maxn],pow=new int [maxn];static int cnt;static void resolve(int x){cnt=0;Arrays.fill(pow,0);for(int i=2;i*i<=x;i++){if(x%i==0){fac[cnt]=i;while(x%i==0){x/=i;pow[cnt]++;}cnt++;}}if(x>1){fac[cnt]=x;pow[cnt++]++;}}public static void main(String[] args) {int n,k,sum1,sum2;Scanner sc=new Scanner(System.in);while(sc.hasNextInt()){n=sc.nextInt();if(n==0) break;k=n+1;while(k>n){resolve(k);if(fac[0]==k){k++;continue;}sum2=0;for(int i=0;i<cnt;i++){sum2+=getsum(fac[i])*pow[i];}sum1=getsum(k);if(sum1==sum2) break;k++;}System.out.println(k);}}}