Sliding window problem. You can bail on computing sub-products whenever you encounter a zero...

int max_prod(string num, size_t n, int k) { int res = 0; size_t l = 0; size_t r;

while (l < (n - k)) {
    r = l;
    int tmp = 1;
    while (tmp > 0 && r < l + k) {
        tmp *= (num[r] - '0');
        if ((r - l == (k - 1)) && tmp > res) {
            res = tmp;
        }
        r++;
    }
    l++;
}

return res;

}