hdu--DFS

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("%d\n",i);

22     }

23     return 0;

24 }

View Code