题目链接：Click here
Solution:
先把式子列出来
[sum_{i_1=L} ^{H}sum_{i_2=L}^H dotssum_{i_n=L}^H [gcd(i_{j=1}^n)=k]\
]
接下来就是莫反套路了
[sum_{i_1=lfloor{L-1 over k}
floor} ^{lfloor{Hover k}
floor}sum_{i_2=lfloor{L-1 over k}
floor}^{lfloor{Hover k}
floor} dotssum_{i_n=lfloor{L-1 over k}
floor}^{lfloor{Hover k}
floor} [gcd(i_{j=1}^n)=1]\
sum_{d=1}^{lfloor{Hover k}
floor}mu(d)sum_{i_1=lfloor{L-1 over kd}
floor} ^{lfloor{Hover kd}
floor}sum_{i_2=lfloor{L-1 over kd}
floor}^{lfloor{Hover kd}
floor} dotssum_{i_n=lfloor{L-1 over kd}
floor}^{lfloor{Hover kd}
floor} \
]
把后面的东西提出来，等于((lfloor{Hover kd} floor-lfloor{L-1 over kd} floor)^n)，即
[sum_{d=1}^{lfloor{Hover k}
floor}mu(d)(lfloor{Hover kd}
floor-lfloor{L-1 over kd}
floor)^n
]
杜教筛筛(mu)是基本操作，在数论分块即可
Code:
#include<bits/stdc++.h>
#define int long long
using namespace std;
const int N=2e6+11;
const int mod=1e9+7;
const int inf=1919191919;
bool vis[N];
int L,R,n,k,cnt;
int ans,p[N],u[N];
unordered_map<int,int> vu;
int read(){
int x=0,f=1;char ch=getchar();
while(!isdigit(ch)){if(ch=='-')f=-f;ch=getchar();}
while(isdigit(ch)){x=x*10+ch-48;ch=getchar();}
return x*f;
}
int qpow(int a,int b){
int re=1;
while(b){
if(b&1) re=re*a%mod;
b>>=1;a=a*a%mod;
}return re%mod;
}
void prepare(){
u[1]=1;
for(int i=2;i<N;i++){
if(!vis[i]) p[++cnt]=i,u[i]=-1;
for(int j=1;j<=cnt&&i*p[j]<N;j++){
vis[i*p[j]]=1;
if(i%p[j]==0) break;
u[i*p[j]]=-u[i];
}
}
for(int i=1;i<N;i++) u[i]+=u[i-1];
}
int getS(int x){
if(x<N) return u[x];
if(vu[x]) return vu[x];
int re=1;
for(int i=2,j;i<=x;i=j+1){
j=x/(x/i);
re-=getS(x/i)*(j-i+1)%mod;
re=(re+mod)%mod;
}return vu[x]=re;
}
signed main(){
prepare();
n=read(),k=read(),L=read(),R=read();
--L;L/=k;R/=k;
for(int i=1,j;i<=R;i=j+1){
j=min(L/i?L/(L/i):inf,R/(R/i));
int v=(getS(j)-getS(i-1))%mod;
int tmp=(R/i-L/i);tmp=qpow(tmp,n);
ans+=tmp*v%mod;ans=(ans+mod)%mod;
}
printf("%lld
",ans);
return 0;
}