993-product of digits

Question: 993-product of digits

A number is converted into the smallest number composed of factors by means of factorization.

Solution: to form a number, the factor only needs to start from 2 to 9, and because the minimum number is required, the factorization starts from the maximum factor, so the number of digits is the least, finally, the output is output from the smallest factor to ensure the minimum. 1 requires special determination, because all numbers have a factor of 1.

Code:

#include <stdio.h>#include <string.h>const int N = 10;int c[N];int n;bool factor() {memset (c, 0, sizeof (c));for (int i = N - 1; i >= 2; i--) {if (n == 1)return true;while (n % i == 0) {c[i]++;n /= i;}}if (n == 1)return true;return false;}int main () {int t;int c6;scanf ("%d", &t);while (t--) {scanf ("%d", &n);if (n != 1) {if (!factor())printf ("-1\n");else {for (int i = 2; i < 10 ; i++) for (int j = 0; j < c[i]; j++)printf ("%d", i);printf ("\n");}} else printf ("1\n");}return 0;}