题目见 http://pan.baidu.com/s/1o6zajc2

此外不知道H-L<=10^5这个条件是干嘛的。。。。

#include <map>
#include <cstdio>
#include <cstring>
#include <iostream>
#include <algorithm>
#define M 10001000
#define INF 0x3f3f3f3f
#define MOD 1000000007
using namespace std;
int mu[M],prime[1001001],tot;
bool not_prime[M];
map<int,long long> mu_sum;
long long n,d,l,r;
void Linear_Shaker()
{int i,j;mu[1]=1;for(i=2;i<=10000000;i++){if(!not_prime[i]){mu[i]=-1;prime[++tot]=i;}for(j=1;prime[j]*i<=10000000;j++){not_prime[prime[j]*i]=true;if(i%prime[j]==0){mu[prime[j]*i]=0;break;}mu[prime[j]*i]=-mu[i];}}for(i=1;i<=10000000;i++)mu[i]+=mu[i-1];
}
long long Mu_Sum(int x)
{if(x<=10000000)return mu[x];if(mu_sum.find(x)!=mu_sum.end())return mu_sum[x];long long i,last,re=1;for(i=1;i<=x;i=last+1){last=x/(x/i);if(x/i-1)re-=(Mu_Sum(last)-Mu_Sum(i-1))*(x/i-1);}return mu_sum[x]=re;
}
long long Quick_Power(long long x,int y)
{long long re=1;while(y){if(y&1) (re*=x)%=MOD;(x*=x)%=MOD; y>>=1;}return re;
}
long long Solve()
{long long i,last,re=0;for(i=1;i<=r;i=last+1){last=min(r/(r/i),l/i?(l/(l/i)):INF);re+=(Mu_Sum(last)-Mu_Sum(i-1))*Quick_Power(r/i-l/i,n);re%=MOD;}return (re%MOD+MOD)%MOD;
}
int main()
{cin>>n>>d>>l>>r;l=(l-1)/d;r=r/d;Linear_Shaker();cout<<Solve()<<endl;
}