UVa 993: Product of digits

这道题很简单。先将N用2,3,5,7（即10以内的素数）分解因数（需要先特殊判断N不为1），然后将可以合并的因数合并（如2*2合并成4，）这样求得的结果位数会减少，大小肯定会小一些。具体实现见代码。

我的解题代码如下：

#include <iostream>
#include <cstdio>
#include <cstring>
#include <cmath>
#include <cstdlib>
#include <string>
#include <algorithm>
using namespace std;

int c[12];
int T;
int N;

int main()
{
cin >> T;
while(T--)
{
cin >> N;
if(N==1) { cout << 1 << endl; continue; }
memset(c,0,sizeof(c));
while(N!=1)	//分解N
{
if(N%2==0) { c[2]++; N/=2; }
else if(N%3==0) { c[3]++; N/=3; }
else if(N%5==0) { c[5]++; N/=5; }
else if(N%7==0) { c[7]++; N/=7; }
else break;
}
if(N!=1) { cout << -1 << endl; continue; }
while(1)	//合并N的因子
{
if(c[2]>=3) { c[2]-=3; c[8]++; }	//因子有三个2，合并为8
else if(c[2]>=2) { c[2]-=2; c[4]++; }	//有两个2，合并为4
else if(c[3]>=2) { c[3]-=2; c[9]++; }	//有两个3，合并为9
else if(c[2]>=1 && c[3]>=1) { c[2]--; c[3]--; c[6]++; }	//有一个2和一个3，合并为6
else break;
}
for(int i=2; i<=9; i++)
{//输出结果
while(c[i])
{
cout << i;
c[i]--;
}
}
cout << endl;
}
return 0;
}

附上题目如下：

For a given non-negative integer number N , find the minimal natural Q such that the product of all digits of Q is equal N .

Input

The first line of input contains one positive integer number, which is the number of data sets. Each subsequent line contains one data set which consists of one non-negative integer number N (0

N

109) .

Output

For each data set, write one line containing the corresponding natural number Q or `-1' if Q does not exist.

Sample Input

3
1
10
123456789

Sample Output

1
25
-1