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 ……

分析：9的阶乘为362880， 9！*10 而且由0~9的阶乘组成的最大数就是3628800。

而且0的阶乘是1，而不是0. 因为根据阶乘定义 n！=n*（n-1）！； 1！=1*0！=1； 所以人为规定了0！=1；

#include 
     
      #include 
      
       #include 
       
        int k[10]= {
    1,1};void ff(){    int i;    for(i = 2; i < 10; i ++){        k[i] = k[i-1]*i;      //  printf("%d\n",k[i]);    }}int main(){    ff();    long i;    long a,sum;    for(i=1; i<=3628800; i++)    {        a=i;        sum=0;        while(a!=0)//a>0        {            sum+=k[a%10];            a=a/10;            //printf("%d\n",a);        }        if(sum==i)            printf("%ld\n",i);    }    return 0;}