正题

P3172

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题目大意

在 [L,R] 选n个数，问gcd=k的方案数

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解题思路

因为gcd=k，那么所选的数都是k的倍数，那么可以让L,R整除k，那么有

∑ a 1 = L R ∑ a 2 = L R . . . ∑ a n = L R [ g c d ( a 1 , a 2 . . . a n ) = 1 ] \sum_{a_1=L}^R\sum_{a_2=L}^R...\sum_{a_n=L}^R[gcd(a_1,a_2...a_n)=1] a1​=L∑R​a2​=L∑R​...an​=L∑R​[gcd(a1​,a2​...an​)=1]

∑ a 1 = l r ∑ a 2 = l r . . . ∑ a n = l r ∑ d ∣ a 1 , d ∣ a 2 . . . d ∣ a n μ ( d ) \sum_{a_1=l}^r\sum_{a_2=l}^r...\sum_{a_n=l}^r\sum_{d|a_1,d|a_2...d|a_n}\mu(d) a1​=l∑r​a2​=l∑r​...an​=l∑r​d∣a1​,d∣a2​...d∣an​∑​μ(d)

∑ d = 1 n μ ( d ) ( r d − l − 1 d ) n \sum_{d=1}^n\mu(d)(\frac{r}{d}- \frac{l-1}{d})^n d=1∑n​μ(d)(dr​−dl−1​)n

然后可以整除分块， μ \mu μ 用杜教筛求

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code

#include<map>
#include<cstdio>
#include<cstring>
#include<iostream>
#include<algorithm>
#define ll long long
#define N 2000210
#define mod 1000000007
using namespace std;
ll n,k,L,R,w,ans,mu[N],p[N],prime[N];
map<ll,ll>smu;
const ll MX=2e6;
void work()
{
	mu[1]=1;
	for(ll i=2;i<=MX;++i){
		if(!p[i]){
			mu[i]=-1;
			prime[++w]=i;
		}
		for(ll j=1;j<=w&&i*prime[j]<=MX;++j){
			p[i*prime[j]]=1;
			if(i%prime[j]==0)break;
			mu[i*prime[j]]=-mu[i];
		}
	}
	for(ll i=2;i<=MX;++i)
		mu[i]+=mu[i-1];
	return;
}
ll S(ll n)
{
	if(n<=MX)return mu[n];
	if(smu.find(n)!=smu.end())return smu[n];
	ll g=1;
	for(ll l=2,r;l<=n;l=r+1){
		r=n/(n/l);
		g-=S(n/l)*(r-l+1);
	}
	smu[n]=g;
	return g;
}
ll ksm(ll x,ll y)
{
	ll z=1;
	while(y){
		if(y&1)z=z*x%mod;
		x=x*x%mod;
		y>>=1;
	}
	return z;
}
int main()
{
	work();
	scanf("%lld%lld%lld%lld",&n,&k,&L,&R);
	L=(L+k-1)/k-1;
	R/=k;
	for(ll l=1,r;l<=R;l=r+1){
		if(L/l)r=min(L/(L/l),R/(R/l));
		else r=R/(R/l);
		(ans+=(S(r)-S(l-1))*ksm(R/l-L/l,n)%mod+mod)%=mod;
	}
	printf("%lld",ans);
	return 0;
}