993 – Product of digits

A factorization problem.
A straight-forward solution would be dividing the input by the numbers from 9 down to 2 and build the output. If after these divisions the input has value other than 1, then it can’t be solved (there are factors larger than 9).
C++ implementation:

#include <cstdio>
#include <cstring>
#include <algorithm>
using namespace std;

typedef unsigned long long int64;

int n;
int64 q, d;

int main() {
	int nCase;
	scanf("%d", &nCase);
	while (nCase--) {
		// in
		scanf("%d", &n);
		if (n == 0) {
			printf("10\n");
			continue;
		} else if (n < 10) {
			printf("%d\n", n);
			continue;
		}
		d = (q = 0) + 1;
		// solve
		for (int i = 9; i > 1; --i)
			while (n % i == 0) {
				n /= i;
				q += i * d;
				d *= 10;
			}
		// out
		if (n == 1)
			printf("%lld\n", q);
		else
			printf("-1\n");
	}
	return 0;
}