Given a positive integer a, find the smallest positive integer b whose multiplication of each digit equals to a.

If there is no answer or the answer is not fit in 32-bit signed integer, then return 0.

Example 1
Input:

48 

Output:

68

Example 2
Input:

15

Output:

35

想法：分解时从较大的因子9-2，不断分解，直至等于1或者不可分。

class Solution {
public:
   vector<int> as;
    int smallestFactorization(int a) {
        if(a<10) return a;
        bool r=factor(a);
        if(r==false || as.size()>32)
        return 0;
        long ans=0;
        for(int i=0;i<as.size();i++){
            ans+=(long)pow(10,i)*as[i];
        }
        return ans>INT_MAX?0:ans;
    }
    bool factor(int a){
       
        while(a>1){
            int i;
            for(i=9;i>1;i--){
                if(a%i==0){
                    a=a/i;
                    as.push_back(i);
                    break;
                }
            }
            if(i==1){
                return false;
            }
        }
        return true;;
    }
};