PAT_A1096#Consecutive Factors

Source:

PAT A1096 Consecutive Factors (20 分)

Description:

Among all the factors of a positive integer N, there may exist several consecutive numbers. For example, 630 can be factored as 3×5×6×7, where 5, 6, and 7 are the three consecutive numbers. Now given any positive N, you are supposed to find the maximum number of consecutive factors, and list the smallest sequence of the consecutive factors.

Input Specification:

Each input file contains one test case, which gives the integer N (1<N<).

Output Specification:

For each test case, print in the first line the maximum number of consecutive factors. Then in the second line, print the smallest sequence of the consecutive factors in the format factor[1]*factor[2]*...*factor[k], where the factors are listed in increasing order, and 1 is NOT included.

Sample Input:

630

Sample Output:

3
5*6*7

Keys:

• 质因子

Attention:

• 注意质数的情况
• 若n存在因子factor0，使得factor0*factor0>n，则factor0唯一；即其余factor，全部有factor*factor<=n

Code:

 /*
 Data: 2019-07-08 19:35:10
 Problem: PAT_A1096#Consecutive Factors
 AC: 31:02

 题目大意：
 正整数N可以分解为多组因数乘积的形式，选择含有最长连续因数的一组，打印该连续因数
 因数中不含1，打印序列递增且较小的
 */
 #include<cstdio>
 #include<vector>
 #include<cmath>
 using namespace std;
 typedef long long LL;

 int main()
 {
     LL n;
     scanf("%lld", &n);
     vector<LL> ans,temp;
     for(LL i=; i<=(LL)sqrt(1.0*n); i++)
     {
         LL sum=;
         temp.clear();
         for(LL j=i; j<=n; j++)
         {
             sum *= j;
             if(n%sum != )
                 break;
             temp.push_back(j);
         }
         if(temp.size() > ans.size())
             ans = temp;
     }
     if(ans.size()==)
         printf("1\n%lld", n);
     else
         printf("%d\n", ans.size());
     for(int i=; i<ans.size(); i++)
         printf("%d%c", ans[i], i==ans.size()-?'\n':'*');

     return ;
 }