hdu 2212 DFS

DFS

Time Limit: 5000/2000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 2325    Accepted Submission(s): 1395

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
......
   
   
   
   
   
   
   
   

    
    
    
    《1》
   
   
   
   
#include <stdio.h>  
int main() 
{ 
 printf("1\n2\n145\n40585\n"); 
 return 0; 
}
《2》
#include<iostream> using namespace std; int fac(int n); int main() {    int i,j,n,m,t,f;    for(i=1;i<=40585;i++)    {       t=i; f=0;                          while(t)       {         n=t%10;         m=t/10;         t=m;         f+=fac(n);       }      if(f==i) cout<<i<<endl;                          }        } int fac(int n) {    int i,ans=1;    for(i=1;i<=n;i++)    ans*=i;    return ans;   }