【oj每周推荐】（算法）各位乘积,该怎么解决

【oj每周推荐】（算法）各位乘积
给出整数N(0 ≤ N ≤ 10^9)，找出一个最小的整数Q，使得将Q的每一位相乘之后等于N
例如N=18，则Q可能取值为：29(2×9=18),36(3×6=18),63(6×3=18),92(9×2=18)
那么我们只要取最小值29即为结果
输入：整数N(0 ≤ N ≤ 10^9)
输出：如果存在这样的Q，则输出Q，如果不存在，输出-1

[解决办法]
string sTemp = "";
int temp = N;
int count2 = two(ref temp);

int count3 = three(ref temp);

int count5 = five(ref temp);
int count7 = seven(ref temp);
if (temp > 9)
{
Q = "-1";
return;
}
int count8 = 0;
int count4 = 0;
int count9 = 0;
int count6 = 0;
while (count3 >= 2)
{
count9++;
count3 = count3 - 2;
}

if (count3 == 1 && count2 > 0)
{
count3 = 0;
count2 = count2 - 1;
count6 = 1;
}

while (count2 >= 3)
{
count8++;
count2 = count2 - 3;
}
if (count2 == 2)
{
count4 = 1;
count2 = 0;
}
else if (count2 == 1)
{
count2 = 1;
}
else
{
count2 = 0;
}


for (int i = 0; i < count2; i++)
{
sTemp = sTemp + "2";
}
for (int i = 0; i < count3; i++)
{
sTemp = sTemp + "3";
}
for (int i = 0; i < count4; i++)
{
sTemp = sTemp + "4";
}
for (int i = 0; i < count5; i++)
{
sTemp = sTemp + "5";
}
for (int i = 0; i < count6; i++)
{
sTemp = sTemp + "6";
}
for (int i = 0; i < count7; i++)
{
sTemp = sTemp + "7";
}
for (int i = 0; i < count8; i++)
{
sTemp = sTemp + "8";
}
for (int i = 0; i < count9; i++)
{
sTemp = sTemp + "9";
}

Q = int.paser（sTemp ）；
[解决办法]
private long t(long N)
{
long r = 0;
for (long m = 1, i = 9; i > 1; i--)
{
while ((N % i) == 0)
{
r += m * i;
m *= 10;
N /= i;
}
}
return r == 0 && N != 0 ? -1 : r;
}
[解决办法]
22楼已经说到了好的方法，我总结下，再可以优化下，但第一次是必须使用9到2分别试除一遍的，不然没法取最小整数。

整理后的代码如下：

C# code

        public static void test(int n)        {            if (n < 0) Console.WriteLine(-1);            else            {                List<int> num = new List<int>();                List<int> q = new List<int>();                //从大到小初始化列表num                for (int i = 9; i > 1; i--)                {                    num.Add(i);                }                //从最大的数开始查找是否可以整除，排除不可整除的，记录可以整除的。                  while (num.Count > 0)                {                    if (n % num[0] == 0)                    {                        n = n / num[0];                        q.Add(num[0]);                    }                    else                        num.Remove(num[0]);                }                //如果最后n为1，说明完全被整除，对记录的各个除数列表排序后输出，得到最小整数，否则输出－1，没有找到                  if (n == 1)                {                    q.Sort();                    foreach (int i in q)                    {                        Console.Write(i);                    }                    Console.WriteLine();                }                else                    Console.WriteLine(-1);            }        } 
 
[解决办法]
自己写了个函数不知道可不可以哈
C# code
private int getV(int n)        {            if (n < 10)                return n;            int temp = n;            List<int> LS = new List<int>();            for (int i = 9; i > 2; i--)            {                while (temp % i == 0)                {                    temp = temp / i;                    LS.Add(i);                }            }            if (temp > 1)                return -1;            else            {                int rv=0;                for (int j = 0; j < LS.Count; j++)                {                    rv += LS[j] * (int)System.Math.Pow(10, j);                }                return rv;            }        }
[解决办法]
Java 版来凑个热闹，呵呵

Java code
public class Test5 {        public static void main(String[] args) {                int n = findMinNumber(100);        System.out.println(n);    }        public static int findMinNumber(int productResult) {        if(productResult < 2) {            return productResult;        }        int result = 0;        for(int i = 9, k = 1; i > 1; i--) {            while(productResult % i == 0) {                result += k * i;                k *= 10;                productResult /= i;            }        }        if(productResult == 1) {            return result;        }        return -1;    }}