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[10]={0,4,27,3125,823543,285311670611,302875106592253};

long long k,r,ans;

vector<long long> v;

void dfs(int x,long long nownum,long long r)
{
    if(x>6)
    {
        if(r)
            return ;
        if(nownum<=k)
        {
            ans++;
            v.push_back(nownum);
        }
        return ;
    }
    long long nn=1;
    for(int i=0;i<=r;i++)
    {
        if(i==0)
            dfs(x+1,nownum,r);
        else
        {
            if(k/nn<num[x])
                break;
            else
                nn*=num[x];
            if(k/nownum<nn)
                break;
            dfs(x+1,nownum*nn,r-i);
        }
    }
    return ;
}

int main()
{
    cin>>r>>k;
    if(r==1)
    {
        for(int i=1;i<=6;i++)
            if(num[i]<=k)
                ans++;
        cout<<ans<<endl;
        for(int i=1;i<=ans;i++)
            cout<<num[i]<<' ';
        return 0;
    }
    dfs(1,1,r);
    cout<<ans<<endl;
    sort(v.begin(),v.end());
    for(int i=0;i<ans;i++)
        cout<<v[i]<<' ';
    return 0;
}