BZOJ 3930: [CQOI2015]选数 莫比乌斯反演 + 杜教筛

求 $\sum_{i=L}^{R}\sum_{i'=L}^{R}....[gcd_{i=1}^{n}(i)==k]$

 

$\Rightarrow \sum_{i=\frac{L}{k}}^{\frac{R}{k}}\sum_{i'=\frac{L}{k}}^{\frac{R}{k}}....[gcd_{i=1}^{n}(i)==1]$

 

$\Rightarrow \sum_{i=\frac{L}{k}}^{\frac{R}{k}}\sum_{i'=\frac{L}{k}}^{\frac{R}{k}}....\sum_{d|gcd_{i=1}^{n}(i)}\mu(d)$

 

$\Rightarrow\sum_{d=1}^{\frac{R}{d}}\mu(d)(\left \lfloor \frac{R}{kd} \right \rfloor-\left \lfloor \frac{L-1}{kd} \right \rfloor)^n$

 

用杜教筛算莫比乌斯函数前缀和，整除分块算一下就行.

#include<bits/stdc++.h>
#define maxn 1040000
#define M 1000001
#define inf 0x7f7f7f7f
#define ll long long
using namespace std;
ll mod = 1000000007;
void setIO(string s)
{
	string in=s+".in";
	freopen(in.c_str(),"r",stdin);
}
map<int,ll>ansmu;
int cnt;
bool vis[maxn];
int prime[maxn], mu[maxn];
ll sumv[maxn];
ll qpow(ll base,ll k)
{
	ll tmp=1;
	while(k)
	{
		if(k&1) tmp=tmp*base%mod;
		base=base*base%mod;
		k>>=1;
	}
	return tmp;
}
void Linear_shaker()
{
	mu[1]=1;
	int i,j;
	for(i=2;i<=M;++i)
	{
		if(!vis[i]) prime[++cnt]=i, mu[i]=-1;
		for(j=1;j<=cnt&&1ll*i*prime[j]<=M;++j)
		{
			vis[i*prime[j]]=1;
			if(i%prime[j]==0)
			{
				mu[i*prime[j]]=0;
				break;
			}
			mu[i*prime[j]]=-mu[i];
		}
	}
	for(i=1;i<=M;++i) sumv[i]=(sumv[i-1]+mu[i]+mod)%mod;
}
ll get(ll n)
{
	if(n<=M) return sumv[n];
	if(ansmu[n]) return ansmu[n];
	ll i,j,re=0;
	for(i=2;i<=n;i=j+1)
	{
		j=(n/(n/i));
		re=(re+(j-i+1)%mod*get(n/i)%mod+mod)%mod;
	}
	return ansmu[n]=(1ll-re+mod)%mod;
}
int main()
{
	// setIO("input");
	ll n,k,L,R,i,j,re=0;
	scanf("%lld%lld%lld%lld",&n,&k,&L,&R);
	L = (L - 1) / k, R = R / k;
	Linear_shaker();
	for(i=1;i<=R;i=j+1)
	{
		j=min(R/(R/i), L/i?L/(L/i):inf);
		re=(re+qpow(R/i-L/i, n) * (get(j)-get(i-1)+mod)%mod)%mod;
	}
	printf("%lld\n",re);
	return 0;
}