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  • 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$ le$N$ le$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
    

     


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  • 原文地址:https://www.cnblogs.com/dyllove98/p/3228631.html
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