XVII Open Cup named after E.V. Pankratiev Stage 14, Grand Prix of Tatarstan, Sunday, April 2, 2017 Problem A. Arithmetic Derivative

题目：Problem A. Arithmetic Derivative
Input file: standard input
Output file: standard input
Time limit: 1 second
Memory limit: 256 mebibytes
Lets define an arithmetic derivative:
• if p = 1 then p0 = 0;
• if p is prime then p0 = 1;
• if p is not prime then n0 = (a · b)0 = a0 · b + a · b0.
For example, 60 = (2 · 3)0 = 20 · 3 + 2 · 30 = 5.
Given positive integers k and r, find all positive integers n such as n ≤ r and n0 = k · n.
Input
Input contains two integers k and r (1 ≤ k ≤ 30; 1 ≤ r ≤ 2 · 1018).
Output
If there are no such numbers, print 0. Otherwise in first line print m — number of positive integers n does
not exceeding r, for which n0 = k · n. Second line then must contain those m integers in ascending order.
Examples

standard input	standard input
1 100	2 4 27
1 2	0

 #include<iostream>
 #include <vector>
 #include <algorithm>
 using namespace std;

 long long num[]={,,,,,,};

 long long k,r,ans;

 vector<long long> v;

 void dfs(int x,long long nownum,long long r)
 {
     if(x>)
     {
         if(r)
             return ;
         if(nownum<=k)
         {
             ans++;
             v.push_back(nownum);
         }
         return ;
     }
     long long nn=;
     for(int i=;i<=r;i++)
     {
         if(i==)
             dfs(x+,nownum,r);
         else
         {
             if(k/nn<num[x])
                 break;
             else
                 nn*=num[x];
             if(k/nownum<nn)
                 break;
             dfs(x+,nownum*nn,r-i);
         }
     }
     return ;
 }

 int main()
 {
     cin>>r>>k;
     if(r==)
     {
         for(int i=;i<=;i++)
             if(num[i]<=k)
                 ans++;
         cout<<ans<<endl;
         for(int i=;i<=ans;i++)
             cout<<num[i]<<' ';
         return ;
     }
     dfs(,,r);
     cout<<ans<<endl;
     sort(v.begin(),v.end());
     for(int i=;i<ans;i++)
         cout<<v[i]<<' ';
     return ;
 }