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 <stdio.h>

#include <stdlib.h>

#include <iostream>

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;

}