Hangzhou Electric School Competition (Counter Attack Index)

逆袭指数

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 1854    Accepted Submission(s): 374

Problem Description

  这依然是关于高富帅小明曾经的故事——   尽管身处逆境,但小明一直没有放弃努力,除了搬砖,小明还研究过东方的八卦以及西方的星座,I have been trying to find the basis for my counterattack in numerology.   When all of these fail,Xiao Ming turned to mathematical research,Hope to get some information from it.一天,Xiao Ming is researching《BestCoderThe mathematical basis of counterattack》这本书时,Found valuable information,其中写道：   Everyone has a counterattack index,For this counterattack index,There may be continuous factors,If this continuous factor is long enough,Then the probability of this person counterattack is very high！   Xiao Ming knew his counterattack index,Please tell Xiao Ming his longest continuous factor,Let him judge whether he can counterattack himself.

 

Input

输入包含多组测试数据. 每组数据占一行,包含一个整数N,Indicates Xiao Ming's counterattack index,N小于2^31.

 

Output

对于每组数据,请输出2行： The first line outputs the longest number of factors; The second line outputs the smallest sequence of factors,Please refer to the example for details. 特别说明：Because Xiao Ming hates being single very much,所以1Not a factor.

 

Sample Input

630

12

 

Sample Output

Hint

630 = 3*5*6*7

 思路：It's hard to look at,After reading the blog of the great god, I realized how much I have,God's address：http://www.1or1.xyz/?p=182

AC代码：

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       #include <iostream>
       
       

       
       #include <cstdio>
       
       

       
       #include <cstdlib>
       
       

       
       #include <algorithm>
       
       

       
       #include <queue>
       
       

       
       #include <stack>
       
       

       
       #include <map>
       
       

       
       #include <cstring>
       
       

       
       #include <climits>
       
       

       
       #include <cmath>
       
       

       
       #include <cctype>
       
       

       
       

       
       typedef 
       
       long 
       
       long 
       
       ll;
       
       

       
       using 
       
       namespace 
       
       std;
       
       

       
       

       
       int 
       
       n;
       
       

       
       

       
       int 
       
       cf(
       
       int 
       
       l)
       
       
{
       
       
    
       
       int 
       
       sum 
       
       = 
       
       0;
       
       
    
       
       int 
       
       temp 
       
       = 
       
       n;
       
       //whilewhich cannot be directly divided byn,will lead to latern值不对
       
       
    
       
       while(
       
       temp)
       
       
    {
       
       
        
       
       if(
       
       temp 
       
       % 
       
       l 
       
       == 
       
       0)
       
       
        {
       
       
            
       
       temp 
       
       = 
       
       temp 
       
       / 
       
       l;
       
       
            
       
       sum
       
       ++;
       
       
            
       
       l
       
       ++;
       
       
        }
       
       
        
       
       else
       
       
            
       
       break;
       
       
    }
       
       
    
       
       return 
       
       sum;
       
       
}
       
       

       
       

       
       int 
       
       main()
       
       
{
       
       
    
       
       int 
       
       i;
       
       
    
       
       while(
       
       scanf(
       
       "%d",
       
       &
       
       n) 
       
       != 
       
       EOF)
       
       
    {
       
       
        
       
       if(
       
       n 
       
       == 
       
       1)
       
       
        {
       
       
            
       
       printf(
       
       "0\n");
       
       
            
       
       continue;
       
       
        }
       
       
        
       
       int 
       
       k;
       
       
        
       
       int 
       
       max1 
       
       = 
       
       0;
       
       
        
       
       int 
       
       sum 
       
       = 
       
       0;
       
       
        
       
       for(
       
       i
       
       =
       
       2; 
       
       i
       
       *
       
       i
       
       <=
       
       n; 
       
       i
       
       ++)
       
       
        {
       
       
            
       
       if(
       
       n 
       
       % 
       
       i 
       
       == 
       
       0)
       
       
            {
       
       
                
       
       sum 
       
       = 
       
       cf(
       
       i);
       
       
            }
       
       
            
       
       if(
       
       sum 
       
       > 
       
       max1)
       
       
            {
       
       
                
       
       max1 
       
       = 
       
       sum;
       
       
                
       
       k 
       
       = 
       
       i;
       
       
            }
       
       
        }
       
       
        
       
       if(
       
       max1 
       
       == 
       
       0)
       
       
        {
       
       
            
       
       printf(
       
       "1\n");
       
       
            
       
       printf(
       
       "%d\n",
       
       n);
       
       
            
       
       continue;
       
       
        }
       
       
        
       
       printf(
       
       "%d\n",
       
       max1);
       
       
        
       
       for(
       
       i
       
       =
       
       0; 
       
       i
       
       <
       
       max1; 
       
       i
       
       ++)
       
       
        {
       
       
            
       
       if(
       
       i 
       
       == 
       
       0)
       
       
                
       
       printf(
       
       "%d",
       
       k);
       
       
            
       
       else
       
       
                
       
       printf(
       
       "*%d",
       
       ++
       
       k);
       
       
        }
       
       
        
       
       printf(
       
       "\n");
       
       
    }
       
       
    
       
       return 
       
       0;
       
       
}
      
      
      
      

       
       
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