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

……

其实你输出后就会知道。。输出就只有4个数，你可以直接输出。

在这里，我是写了过程的。

public class Main{

    static int fact[] = new int[10];

    public static void main(String[] args) {

        dabiao();

        //9!*7 7位数-比9999999小，后面的数字更不用说了，肯定小。

        for(int i=1;i<=9999999;i++){

            if(isTrue(i)){

                System.out.println(i);

            }

        }

    }

    private static void dabiao() {

        //求阶乘的，注意：0的阶乘为1

        fact[0]=1;

        for(int i=1;i<fact.length;i++){

            fact[i]=1;

            for(int j=1;j<=i;j++){

                fact[i]=fact[i]*j;

            }

        }

    }

    //判断是不是相等

    private static boolean isTrue(int i) {

        if(i==1||i==2){

            return true;

        }

        int sum=0;

        int n=i;

        while(n!=0){

            int k=n%10;

            sum+=fact[k];

            n=n/10;

        }

        if(sum==i){

            return true;

        }

        return false;

    }

}