DFS

Time Limit: 5000/2000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others) Total Submission(s): 4422    Accepted Submission(s): 2728

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

......

Author

zjt

水体.....暴搜....

代码：

 1 //hdu 2212 
 2 //@Gxjun coder 
 3 #include<stdio.h>
 4 int main()
 5 {
 6     int val[10]={1,1},i;
 7     //求出1~9的阶乘
 8     for(i=2;i<=9;i++)
 9          val[i]=val[i-1]*i;
10      int ans,tem;
11     for( i=1 ; i<= 3628800 ; i++)
12     {
13         tem=i;
14         ans=0;
15         while(tem)
16         {
17             ans+=val[tem%10];
18             tem/=10;
19         }
20         if(ans==i)
21             printf("%dn",i);
22     }
23     return 0;
24 }