BZOJ-3930 [CQOI2015]选数(莫比乌斯反演+杜教筛)

题目描述

  从区间 \([L,R]\) 中选取 \(N\) 个整数，总共有 \((R-L+1)^{N}\) 种方案。对每种方案选出的 \(N\) 个整数都求一次最大公约数，求最大公约数刚好为 \(K\) 的选取方案有多少个。

  数据范围：\(1\leq N,K\leq 10^{9},1\leq L\leq R\leq 10^{9},R-L\leq 10^{5}\)。

分析

  题目即求：

\[\sum_{a_1=L}^{R}\sum_{a2=L}^{R}\cdots\sum_{a_n=L}^{R}[\gcd(a_1,a_2,\cdots,a_N)=K] \]

  根据套路，所有 \(a_i\) 都除 \(K\)，则：

\[L\leq a_i\leq R\Longrightarrow L\leq a_i'\times K\leq R\Longrightarrow\Big\lceil\frac{L}{K}\Big\rceil\leq a_i'\leq \Big\lfloor\frac{R}{K}\Big\rfloor\Longrightarrow\ \Big\lfloor\frac{L+K-1}{K}\Big\rfloor \leq a_i'\leq \Big\lfloor\frac{R}{K}\Big\rfloor \]

\[\begin{aligned}&\sum_{a_1=\lfloor \frac{L+K-1}{K}\rfloor}^{\lfloor\frac{R}{K}\rfloor}\sum_{a_2=\lfloor \frac{L+K-1}{K}\rfloor}^{\lfloor\frac{R}{K}\rfloor}\cdots\sum_{a_N=\lfloor \frac{L+K-1}{K}\rfloor}^{\lfloor\frac{r}{k}\rfloor}[\gcd(a_1,a_2,\cdots,a_N)=1]\\=&\sum_{a_1=L}^{R}\sum_{a2=L}^{R}\cdots\sum_{a_n=L}^{R}[\gcd(a_1,a_2,\cdots,a_N)=K]\\=&\sum_{a_1=\lfloor \frac{L+K-1}{K}\rfloor}^{\lfloor\frac{R}{K}\rfloor}\sum_{a_2=\lfloor \frac{L+K-1}{K}\rfloor}^{\lfloor\frac{R}{K}\rfloor}\cdots\sum_{a_N=\lfloor \frac{L+K-1}{K}\rfloor}^{\lfloor\frac{R}{K}\rfloor}\sum_{d\mid \gcd(a_1,a_2,\cdots,a_N)}\mu(d)\\=&\sum_{d=1}^{\lfloor\frac{R}{K}\rfloor}\mu(d)\sum_{a_1=\lfloor \frac{L+K-1}{K}\rfloor}^{\lfloor\frac{R}{K}\rfloor}\sum_{a_2=\lfloor \frac{L+K-1}{K}\rfloor}^{\lfloor\frac{R}{K}\rfloor}\cdots\sum_{a_N=\lfloor \frac{L+K-1}{K}\rfloor}^{\lfloor\frac{R}{K}\rfloor}[d\mid \gcd(a_1,a_1,\cdots,a_N)]\\=&\sum_{d=1}^{\lfloor\frac{R}{K}\rfloor}\mu(d)\sum_{a_1=\lfloor \frac{L+K-1}{K}\rfloor}^{\lfloor\frac{R}{K}\rfloor}[d\mid a_1]\sum_{a_2=\lfloor \frac{L+K-1}{K}\rfloor}^{\lfloor\frac{R}{K}\rfloor}[d\mid a_N]\cdots\sum_{a_N=\lfloor \frac{L+K-1}{K}\rfloor}^{\lfloor\frac{R}{K}\rfloor}[d\mid a_N]\\=&\sum_{d=1}^{\lfloor\frac{R}{K}\rfloor}\mu(d)\Big(\Big\lfloor\frac{R}{Kd}\Big\rfloor-\Big\lfloor\frac{L-1}{Kd}\Big\rfloor\Big)^N\end{aligned} \]

  对于 \(\mu\) 的前缀和，根据杜教筛公式：

\[S(n)=g(1)S(n)=\sum\limits_{i=1}^{n}(f\ast g)(i)-\sum\limits_{i=2}^{n}g(i)S\Big(\Big\lfloor\frac{n}{i}\Big\rfloor\Big) \]

  取 \(f=\mu,g=I,h=\mu\ast I=\epsilon\)，则：

\[S(n)=1-\sum_{i=2}^{n}S\Big(\Big\lfloor\frac{n}{i}\Big\rfloor\Big) \]

  杜教筛 \(+\) 数论分块处理即可，时间复杂度 \(O(R^{\frac{2}{3}})\)。

代码

#include<bits/stdc++.h>
using namespace std;
const int N=2e6+10;//n^(2/3)
const int mod=1e9+7;
long long prime[N+10],vis[N+10],mu[N+10],sum[N+10],cnt;
long long quick_pow(long long a,long long b)
{
    long long ans=1;
    while(b)
    {
        if(b&1)
            ans=ans*a%mod;
        a=a*a%mod;
        b>>=1;
    }
    return ans;
}
void init()
{
    mu[1]=1;
    for(int i=2;i<=N;i++)
    {
        if(!vis[i])
        {
            prime[++cnt]=i;
            mu[i]=-1;
        }
        for(int j=1;j<=cnt&&i*prime[j]<=N;j++)
        {
            vis[i*prime[j]]=1;
            if(i%prime[j]==0)
            {
                mu[i*prime[j]]=0;
                break;
            }
            else
                mu[i*prime[j]]=-mu[i];
        }
    }
    for(int i=1;i<=N;i++)
        sum[i]=(sum[i-1]+mu[i])%mod;
}
map<long long,long long> mp;
long long S_mu(long long n)
{
    if(n<N)
        return sum[n];
    if(mp[n])
        return mp[n];
    long long ans=1;
    for(long long l=2,r;l<=n;l=r+1)
    {
        r=n/(n/l);
        ans=((ans-S_mu(n/l)*(r-l+1)%mod)%mod+mod)%mod;
    }
    mp[n]=(ans+mod)%mod;
    return (ans+mod)%mod;
}
long long solve(long long N,long long K,long long L,long long R)
{
    L=(L-1)/K;R=R/K;
    long long ans=0;
    for(long long l=1,r;l<=R;l=r+1)
    {
        if(L/l==0)
            r=R/(R/l);
        else
            r=min(L/(L/l),R/(R/l));
        ans=(ans+(S_mu(r)-S_mu(l-1))*quick_pow((R/l)-(L/l),N)%mod)%mod;
    }
    return (ans+mod)%mod;
}
int main()
{
    init();
    long long N,K,L,R;
    cin>>N>>K>>L>>R;
    cout<<solve(N,K,L,R)<<endl;
    return 0;
}