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Posted 2020-10-01 weeping

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

 1 #include<iostream>
 2 #include <vector>
 3 #include <algorithm>
 4 using namespace std;
 5 
 6 long long num[10]={0,4,27,3125,823543,285311670611,302875106592253};
 7 
 8 long long k,r,ans;
 9 
10 vector<long long> v;
11 
12 void dfs(int x,long long nownum,long long r)
13 {
14     if(x>6)
15     {
16         if(r)
17             return ;
18         if(nownum<=k)
19         {
20             ans++;
21             v.push_back(nownum);
22         }
23         return ;
24     }
25     long long nn=1;
26     for(int i=0;i<=r;i++)
27     {
28         if(i==0)
29             dfs(x+1,nownum,r);
30         else
31         {
32             if(k/nn<num[x])
33                 break;
34             else
35                 nn*=num[x];
36             if(k/nownum<nn)
37                 break;
38             dfs(x+1,nownum*nn,r-i);
39         }
40     }
41     return ;
42 }
43 
44 int main()
45 {
46     cin>>r>>k;
47     if(r==1)
48     {
49         for(int i=1;i<=6;i++)
50             if(num[i]<=k)
51                 ans++;
52         cout<<ans<<endl;
53         for(int i=1;i<=ans;i++)
54             cout<<num[i]<<‘ ‘;
55         return 0;
56     }
57     dfs(1,1,r);
58     cout<<ans<<endl;
59     sort(v.begin(),v.end());
60     for(int i=0;i<ans;i++)
61         cout<<v[i]<<‘ ‘;
62     return 0;
63 }