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[11];

int main()
{
    a[0]=1;
    a[1]=1;
    for(i=2;i<=9;i++) a[i]=a[i-1]*i;

     n=a[9]*10;
     //printf("%d\n",n);
    for(i=1;i<=n;i++)
      {
        int t=i;
        int sum=0;
        while(t>0)
        {
          sum+=a[t%10];
          t=t/10;
        }
        if (sum==i) printf("%d\n",i);
      }
    return 0;
}