[P4450] 双亲数 - 莫比乌斯反演,整除分块

模板题……
\[\sum\limits_{i=1}^a\sum\limits_{j=1}^b[(i,j)=k] = \sum\limits_{i=1}^a\sum\limits_{j=1}^b[k|i][k|j][({i\over k},{j\over k})=1]=\sum\limits_{i=1}^{a\over k}\sum\limits_{j=1}^{b\over k}[(i,j)=1]\]
继续化简
\[\sum\limits_{i=1}^{b\over k}\sum\limits_{j=1}^{d\over k}\sum\limits_{t|(i,j)}\mu(t)=\sum\limits_{i=1}^{b\over k}[t|i]\sum\limits_{j=1}^{d\over k}[t|j]\mu(t)=\sum\limits_{t=1}^{max({b\over k},{d\over k})}{\lfloor{{b\over k}\over t}\rfloor}{\lfloor{{d\over k}\over t}\rfloor}\mu(t)\]
然后上反演整除分块即可

#include <bits/stdc++.h>
using namespace std;
#define int long long
const int N = 1000005;

int pr[N*2],is[N*2],mu[N*2],cnt;

signed main() {
    mu[0]=mu[1]=1; is[1]=1;
    for(int i=2;i<N;i++) {
        if(is[i]==0) {
            pr[++cnt]=i;
            mu[i]=-1;
        }
        for(int j=1; j<=cnt&&pr[j]*i<N; ++j) {
            is[pr[j]*i]=1;
            if(i%pr[j]==0) {
                mu[pr[j]*i]=0;
                break;
            }
            else {
                mu[pr[j]*i]=-mu[i];
            }
        }
    }
    for(int i=1;i<N;i++) mu[i]+=mu[i-1];

    int a,b,d;
    cin>>a>>b>>d;
    a/=d; b/=d;
    int ans = 0;
    int m=min(a,b);
    int l=1,r;
    while(l<=m) {
        r=min(a/(a/l),b/(b/l));
        ans+=(mu[r]-mu[l-1])*(a/l)*(b/l);
        l=r+1;
    }
    cout<<ans<<endl;
}

原文地址：https://www.cnblogs.com/mollnn/p/12250375.html