HDOJ--2212 DFS

Problem Description
  
  
   A DFS(digital factorial sum) number is found by summing the factorial of every digit of a positive integer. 

For example ,consider the positive integer 145 = 1!+4!+5!, so it's a DFS number.

Now you should find out all the DFS numbers in the range of int( [1, 2147483647] ).

There is no input for this problem. Output all the DFS numbers in increasing order. The first 2 lines of the output are shown below.
  

  
   Input
  
  
   no input

  
   Output
  
  
   Output all the DFS number in increasing order. 
 

  
   Sample Output
  
 
     1
2
......
    

     
   
import java.util.Scanner;

public class Main {
   static int fn[]=new int[10];

   public static void main(String[] args) {
      Scanner sc=new Scanner(System.in);
        dashu();
        for(int i=1;i<9999999;i++){
           if(isTrue(i)){
           System.out.println(i);
           }
        }

   }
   public static void dashu(){
      fn[0]=1;
      for(int i=1;i<fn.length;i++){
         fn[i]=1;
         for(int j=1;j<=i;j++){
            fn[i]=fn[i]*j;
         }
      }
   }
   public static boolean isTrue(int i){
      if(i==1||i==2){
         return true;
      }
      int n=i,sum=0;
      while(n!=0){
         int m=n%10;
         sum+=fn[m];
         n=n/10;
      }
      if(sum==i){
         return true;
      }
      return false;
   }
}