HDU 5323 DFS

DFS

Time Limit : 5000/2000ms (Java/Other)   Memory Limit : 65536/32768K (Java/Other)

Total Submission(s) : 61   Accepted Submission(s) : 32

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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

#include <iostream>
#include<cstdio>
using namespace std;

int i,n;
int a[];

int main()
{
    a[]=;
    a[]=;
    for(i=;i<=;i++) a[i]=a[i-]*i;

     n=a[]*;
     //printf("%d\n",n);
    for(i=;i<=n;i++)
      {
        int t=i;
        int sum=;
        while(t>)
        {
          sum+=a[t%];
          t=t/;
        }
        if (sum==i) printf("%d\n",i);
      }
    return ;
}