DFS

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

Problem DescriptionA 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.

 Inputno input OutputOutput all the DFS number in increasing order. Sample Output12…… Authorzjt

水体…..暴搜….

代码：

 //hdu 2212
 //@Gxjun coder
 #include<stdio.h>
 int main()
 {
     int val[]={,},i;
     //求出1~9的阶乘
     for(i=;i<=;i++)
          val[i]=val[i-]*i;
      int ans,tem;
     for( i= ; i<=  ; i++)
     {
         tem=i;
         ans=;
         while(tem)
         {
             ans+=val[tem%];
             tem/=;
         }
         if(ans==i)
             printf("%d\n",i);
     }
     return ;
 }