傳送門

分析

對這個$f(k)$整除分塊，用杜教篩搞出$\mu$的部分然後另一部分快速冪即可

程式碼

#include<iostream>
#include<cstdio>
#include<cstring>
#include<string>
#include<algorithm>
#include<cctype>
#include<cmath>
#include<cstdlib>
#include<queue>
#include<ctime>
#include
 
<vector>
#include<set>
#include<map>
#include<stack>
using namespace std;
const int N = 5e6;
const int mod = 1e9+7;
int p[N+10],mu[N+10];
bool is[N+10];
map<int,int>MU;
inline void init(){
    int i,j,cnt=0;
    mu[1]=1;
    for(i=2;i<=N;i++){
      if(!is[i])p[++cnt]=i,mu[i]=-1
 
;
      for(j=1;j<=cnt,i*p[j]<=N;j++){
          is[p[j]*i]=1;
          if(i%p[j]==0){
            mu[p[j]*i]=0;
            break;
          }
          mu[p[j]*i]=-mu[i];
      }
    }
    for(i=2;i<=N;i++)mu[i]=(mu[i]+mu[i-1]+mod)%mod;
}
inline int go(int x){
    if(x<=N)return mu[x];
    
 
if(MU[x])return MU[x];
    int res=1,le=2,ri;
    for(;le<=x;le=ri+1){
      ri=x/(x/le);
      res=(res-(long long)(ri-le+1)*go(x/le)%mod+mod)%mod;
    }
    return MU[x]=res;
}
inline int pw(int x,int p){
    int res=1;
    while(p){
      if(p&1)res=(long long)res*x%mod;
      x=(long long)x*x%mod;
      p>>=1;
    }
    return res;
}
int main(){
    int n,m,p,k,L,R,le=1,ri,Ans=0;
    scanf("%d%d%d%d",&p,&k,&L,&R);
    n=R/k,m=(L-1)/k;
    init();
    for(;le<=n;le=ri+1){
      if(m/le)ri=min(n/(n/le),m/(m/le));
        else ri=n/(n/le);
      Ans=(Ans+(long long)(go(ri)-go(le-1)+mod)%mod*pw(n/le-m/le,p)%mod)%mod;
    }
    printf("%d\n",Ans);
    return 0;
}