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
……

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
",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
",a);
        }
        if(sum==i)
            printf("%ld
",i);
    }
    return 0;
}