UVa 993: Product of digits

這道題很簡單。先將N用2,3,5,7（即10以內的素數）分解因數（需要先特殊判斷N不為1），然後將可以合並的因數合並（如2*2合並成4，）這樣求得的結果位數會減少，大小肯定會小一些。具體實現見代碼。

我的解題代碼如下：

#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)	//合並N的因子
		{
			if(c[2]>=3) { c[2]-=3; c[8]++; }	//因子有三個2，合並為8
			else if(c[2]>=2) { c[2]-=2; c[4]++; }	//有兩個2，合並為4
			else if(c[3]>=2) { c[3]-=2; c[9]++; }	//有兩個3，合並為9
			else if(c[2]>=1 && c[3]>=1) { c[2]--; c[3]--; c[6]++; }	//有一個2和一個3，合並為6
			else break;
		}
		for(int i=2; i<=9; i++)
		{//輸出結果
			while(c[i])
			{
				cout << i;
				c[i]--;
			}
		}
		cout << endl;
	}
	return 0;
}

附上題目如下：

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 (0N109) .

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