洛谷P1587 [NOI2016] 循环之美

若 \(\frac{a}{b}\) 为纯循环小数，设 \(l\) 为其循环节长度，得：

\[\large \frac{a}{b}\left(k^l-1\right) \in\mathbb Z \Rightarrow k^l=1 \pmod{b} \Rightarrow k \perp b \]

所求即为：

\[\large\begin{aligned} &f\left(n,m,k\right)\\ =&\sum_{i=1}^n\sum_{j=1}^m\left[ i \perp j \right]\left[ j \perp k \right]\\ =&\sum_{i=1}^n\sum_{j=1}^m\left[ i \perp j \right]\sum_{d\mid j \and d\mid k}\mu(d)\\ =&\sum_{d\mid k}\mu(d)\sum_{i=1}^n\sum_{j=1}^{\left\lfloor \frac{m}{d} \right\rfloor}\left[ i \perp jd \right]\\ =&\sum_{d\mid k}\mu(d)\sum_{j=1}^{\left\lfloor \frac{m}{d} \right\rfloor}\sum_{i=1}^n\left[ i \perp j \right]\left[ i \perp d \right]\\ =&\sum_{d\mid k}\mu(d)f\left(\left\lfloor \frac{m}{d} \right\rfloor,n,d\right)\\ \end{aligned} \]

边界为：

\[\large\begin{aligned} f\left(0,m,k\right)&=f\left(n,0,k\right)=0\\ f\left(n,m,1\right)&=\sum_{i=1}^n\sum_{j=1}^m\left[ i \perp j \right]=\sum_{i=1}^n\mu(d)\left\lfloor \frac{n}{d} \right\rfloor\left\lfloor \frac{m}{d} \right\rfloor \end{aligned} \]

递归计算即可，算边界是用杜教筛。复杂度为 \(O(n^{\frac{2}{3}}+\sqrt n \log^2 n)\)。

#include<bits/stdc++.h>
#define maxn 10000010
#define maxm 600010
using namespace std;
typedef long long ll;
template<typename T> inline void read(T &x)
{
    x=0;char c=getchar();bool flag=false;
    while(!isdigit(c)){if(c=='-')flag=true;c=getchar();}
    while(isdigit(c)){x=(x<<1)+(x<<3)+(c^48);c=getchar();}
    if(flag)x=-x;
}
ll n,m,k,tot,all=10000000;
ll p[maxn],mu[maxn],s[maxn];
bool tag[maxn];
map<int,ll> sum;
struct edge
{
    int to,nxt;
    edge(int a=0,int b=0)
    {
        to=a,nxt=b;
    }
}e[maxm];
int head[maxm],edge_cnt;
void add(int from,int to)
{
    e[++edge_cnt]=edge(to,head[from]),head[from]=edge_cnt;
}
void init()
{
    all=min(all,max(n,m)),mu[1]=1;
    for(int i=2;i<=all;++i)
    {
        if(!tag[i]) p[++tot]=i,mu[i]=-1;
        for(int j=1;j<=tot;++j)
        {
            int k=i*p[j];
            if(k>all) break;
            tag[k]=true;
            if(i%p[j]) mu[k]=-mu[i];
            else
            {
                mu[k]=0;
                break;   
            }
        }
    }
    for(int i=1;i<=all;++i) s[i]=s[i-1]+mu[i];
    for(int i=1;i<=k;++i)
        if(mu[i])
            for(int j=i;j<=k;j+=i)
                add(j,i);
}
ll S(int n)
{
    if(n<=all) return s[n];
    if(sum.count(n)) return sum[n];
    ll v=1;
    for(int l=2,r;l<=n;l=r+1) r=n/(n/l),v-=S(n/l)*(r-l+1);
    return sum[n]=v;
}
ll f(int n,int m,int k)
{
    if(!n||!m) return 0;
    ll v=0;
    if(k==1)
    {
        if(n>m) swap(n,m);
        for(int l=1,r;l<=n;l=r+1)
            r=min(n/(n/l),m/(m/l)),v+=(S(r)-S(l-1))*(n/l)*(m/l);
    }
    else
    {
        for(int i=head[k];i;i=e[i].nxt)
        {
            int d=e[i].to;
            v+=mu[d]*f(m/d,n,d);
        }
    }
    return v;
}
int main()
{
    read(n),read(m),read(k),init();
    printf("%lld",f(n,m,k));
    return 0;
}