BZOJ 3930 容斥原理

//By SiriusRen
#include <cstdio>
#include <cstring>
#include <algorithm>
using namespace std;
#define int long long
const int mod=1000000007;
int n,k,l,h,f[100050];
int pow(int x,int y){
    int res=1;
    while(y){
        if(y&1)res=res*x%mod;
        x=x*x%mod,y>>=1;
    }return res;
}
signed main(){
    scanf("%lld%lld%lld%lld",&n,&k,&l,&h);
    for(int i=h-l;i;i--){
        int L=(l-1)/(k*i),R=h/(k*i);
        f[i]+=(pow(R-L,n)-R+L+mod)%mod;
        for(int j=2;i*j<=h-l;j++)(f[i]-=f[i*j])%=mod;
        f[i]=(f[i]+mod)%mod;
    }printf("%lld\n",(f[1]+(k>=l&&k<=h))%mod);
}