BZOJ 3930: [CQOI2015]选数 莫比乌斯反演 + 杜教筛

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

 

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

 

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

 

$Rightarrowsum_{d=1}^{frac{R}{d}}mu(d)(left lfloor frac{R}{kd} ight floor-left lfloor frac{L-1}{kd} ight floor)^n$

 

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
",re); 
	return 0; 
}