省选专练CQOI2015选数

考虑H-L很小于是可以暴力枚举H-L中k的倍数

当然，莫比乌斯反演后使用map离散化进行杜教筛也可以过，甚至还可以无视H-L的条件

#include<iostream>
#include<cstdio>
#include<algorithm> 
#include<cmath>
#include<cstring>
using namespace std;
#define ll long long
ll n,k,l,r; 
ll flag=0;
const ll mod=1e9+7;
ll quick_pow(ll x,ll k){
	ll ret=1;
	while(k){
		if(k%2==1){
			ret=(ret*x)%mod;
		}
		x=(x*x)%mod;
		k/=2;
	}
	return ret;
}
ll f[400000]={0};
int main(){
//	freopen("number.in","r",stdin);
//	freopen("number.out","w",stdout); 
	cin>>n>>k>>l>>r;
	if(l<=k&&k<=r){
		flag=1;
	}
	l=(l-1)/k;
	r=r/k; 
	ll Mx=r-l;
	for(int i=Mx;i>=1;i--){
		ll now=i;
		ll L=l/now;
		ll R=r/now;
		if(L<R){
			f[i]=((quick_pow((R-L),n)-(R-L))+mod)%mod;// delete choose all of itself because the GCD is itself
			for(int j=(i<<1);j<=Mx;j+=i){
				f[i]=((f[i]-f[j])+mod)%mod;
			}
		}
	}
//	for(int i=1;i<=Mx;i++){
//		cout<<f[i]<<" ";
//	}
	cout<<f[1]+flag;
	return 0;
}

转载于:https://www.cnblogs.com/Leo-JAM/p/10079286.html