HDu2212

DFS


Time Limit: 5000/2000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 6792    Accepted Submission(s): 4180




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

......

方法一；

       其实大多数的都是通过打表来实现的，这里我也讲一种！其实对付这种题目，最好的方法也就是打表！   

   附代码 

     

       

#include<stdio.h>  int main()  {      printf("1\n2\n145\n40585\n");      return 0;  }  

  

  附打表代码

    

#include<stdio.h>int main(){     __int64 i,j,k,l,a[10],s;a[0]=1;for(i=1;i<=10;i++)a[i]=a[i-1]*i;for(i=1;i<=2147483647;i++)   {    k=i;s=0;      while(k)      {      j=k%10;      k=k/10;      s=s+a[j];  }  if(s==i)   printf("%I64d\n",i);   }return 0;}

   方法二

   其实方法二和方法一差不多，就是去求最大的一个数是多少！我从9999999开始的，然后就8888888…………往前面求，当我求到999999时我发现可以作为上界了；所以……

   附代码  

       

#include<stdio.h>int main(){     __int64 i,j,k,l,a[10],s;a[0]=1;for(i=1;i<=10;i++)a[i]=a[i-1]*i;      for(i=1;i<=999999;i++)      {      k=i;s=0;      while(k)      {      j=k%10;      k=k/10;      s=s+a[j];  }  if(s==i)  printf("%I64d\n",i);  }return 0; }