Competitive Programming #3: [Fact Digit Sum]

Here is the problem: 

Let's define A(n) for positive integer n as a sum of factorials of its digits. For example, A(154)=1!+ 5! +4!= 145.
Given a number 'x' , you need to print the minimum number L such that A(L)=x. If no such L exists, then print -1
Input:
First line contains 'T' - the number of test cases
Then each line contains an integer 'x'.
Output:
Print the answer for each value 'x'.
Constraints:
1<=T<=542
1000<=x<=1000000000
Example:
Input:
1
40321
Output:
18
Explanations:
A(18)=1!+ 8! =40321  and 18 is the smallest element for which A(18) is 40321
Note that A(80) = A(81) is also 40321, But among them 18 is the smallest number.

Solution:
Another interesting problem. I thought it is hard problem but actually it is very easy.(Dont worry I am not good at all with competitive programming  :D )
The approach is with Greedy Algorithm ... So we have:


It is with recursive function , if u still dont understand recursion , u can use 'for'
factoriels[9] = {1! , 2!, 3! , 4! , 5! , 6! , 7! , 8! , 9!}
Example:
1 
5
RecursiveF(5,factoriels[9], 8 , facts);
   ->else
         -> if(factoriels[8] > 5) ? YES #RecursiveF(5,factoriels[9],7,facts) ; 
(so now repeat the step until factioriel[i] < = 5)
        (after that) ->else #facts.push_back(2); #RecursiveF(2,factoriels[9],1,facts) ; 
(repeat until n!=0, when n==0 , it gonna output  the reserve order of elements in array facts)