Number theory can only for loop (mathematics + block + memoization)

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The main idea of ​​the topic:
Given $n,k$ 计算函数：
$$f(n,k)=\{\begin{array}{ll}1& ifk=0\\ {\sum }_{i=1}^n\lceil lo{g}_{2}i\rceil f(\lfloor \frac{n}{i}\rfloor ,k-1)& ifk\ge 1\end{array}$$
The answer is $1e9+7$ modulo

Problem analysis:
Analyzing this formula, it is obvious that $f(1,k)=0(k\ge 1)$
Find$f(\lfloor \frac{n}{i}\rfloor ,k-1)$ medium-specific$\lfloor \frac{n}{i}\rfloor$ is a typical block type, which can be processed in blocks, and$\lfloor \frac{n}{i}\rfloor =1$ This block size is$\frac{n}{2}$, Using this property we can get $f(n,k)$ in$f(1,x)$ is approximately$lo{g}_{2}n$ (harmonic series)
so for$n\ge lo{g}_2(1e8)\ge$The answer to the data of$2$$7$ is$0$
for$n\le lo{g}_2\left(1e8\right)$ we can enumerate$\lceil lo{g}_{2}i\rceil$ then use block processing$f(\lfloor \frac{n}{i}\rfloor ,k-1)$ , just memorize + recursive solution

See the code for details:

#include<iostream>
#include<cstdio>
#include<cstring>
#include<algorithm>
#include<cmath>
#include<vector>
#include<set>
#include<map>
#include<queue>
#include<unordered_map>
#define ll long long
#define inf 0x3f3f3f3f
using namespace std;
int read()
{
    
    
	int res = 0,flag = 1;
	char ch = getchar();
	while(ch<'0' || ch>'9')
	{
    
    
		if(ch == '-') flag = -1;
		ch = getchar();
	}
	while(ch>='0' && ch<='9')
	{
    
    
		res = (res<<3)+(res<<1)+(ch^48);//res*10+ch-'0';
		ch = getchar();
	}
	return res*flag;
}
const int maxn = 5e3+5;
const int mod = 1e9+7;
const double pi = acos(-1);
const double eps = 1e-8;
unordered_map<int,int>mp[27];
int f(int n,int k)
{
    
    
	if(k >= 27) return 0;
	if(mp[k].find(n) != mp[k].end())
		return mp[k][n];
	if(k == 0) return mp[k][n] = 1;
	else {
    
    
		int& tmp = mp[k][n];
		for(int i = 1;(1<<i) <= 2*n;i++)
		{
    
    
			int l = (1<<(i-1))+1,r = min(1<<i,n),rr;
			for(int j = l;j <= r;j = rr+1)
			{
    
    
				int pos = n/j;
				rr = min(r,n/pos);
				tmp = (tmp+1ll*i*(rr-j+1)%mod*f(pos,k-1))%mod;
			}
		}
		return tmp;
	}
}
int main()
{
    
    
	int n = read(),k = read();
	printf("%d\n",f(n,k));
	return 0;
}