P3327 [SDOI2015]约数个数和

思路

做这题先要知道一个性质，

\[d_{ij}=\sum_{x|i}\sum_{y|j}[(x,y)=1]
\]

然后上莫比乌斯反演颓柿子就好了

\[\begin{align}&\sum_{i=1}^n\sum_{j=1}^m\sum_{x|i}\sum_{y|j}[(x,y)=1]\\=&\sum_{i=1}^n\sum_{j=1}^m\sum_{x|i}\sum_{y|j}\sum_{d|(x,y)}\mu(d)\\=&\sum_{i=1}^n\sum_{j=1}^m\sum_d^{min(n,m)}\mu(d)\sum_{x=1}^{\lfloor\frac{i}{d}\rfloor}\sum_{y=1}^{\lfloor\frac{j}{d}\rfloor}1\\=&\sum_d^{min(n,m)}\mu(d)\sum_{i=1}^n\sum_{x=1}^{\lfloor\frac{i}{d}\rfloor}\sum_{j=1}^m\sum_{y=1}^{\lfloor\frac{j}{d}\rfloor}1\\=&\sum_d^{min(n,m)}\mu(d)\sum_{x=1}^{\lfloor\frac{n}{d}\rfloor}\lfloor\frac{n}{xd}\rfloor\sum_{y=1}^{\lfloor\frac{m}{d}\rfloor}\lfloor\frac{m}{dy}\rfloor\\=&\sum_d^{min(n,m)}\mu(d)\sum_{x=1}^{\lfloor\frac{n}{d}\rfloor}\lfloor\frac{\lfloor \frac{n}{d}\rfloor}{x}\rfloor\sum_{y=1}^{\lfloor\frac{m}{d}\rfloor}\lfloor\frac{\lfloor \frac{m}{d} \rfloor}{y}\rfloor\end{align}\\
\]

后面的部分，预处理一个\(\sum_{i=1}^n \lfloor \frac{n}{i} \rfloor\)就好了，前面上一个整除分块

代码

#include <cstring>
#include <cstdio>
#include <algorithm>
using namespace std;
int isprime[50010],mu[50010],cnt,iprime[50010],n,m,t;
long long sum[50010];
void prime(int n){
    mu[1]=1;
    isprime[1]=true;
    for(int i=2;i<=n;i++){
        if(!isprime[i]){
            iprime[++cnt]=i;
            mu[i]=-1;
        }
        for(int j=1;j<=cnt&&iprime[j]*i<=n;j++){
            isprime[i*iprime[j]]=true;
            mu[i*iprime[j]]=-mu[i];
            if(i%iprime[j]==0){
                mu[i*iprime[j]]=0;
                break;
            }
        }
    }
    for(int i=1;i<=n;i++)
        mu[i]+=mu[i-1];
}
void pre(int n){
    for(int i=1;i<=n;i++){
        long long ans=0;
        for(int l=1,r;l<=i;l=r+1){
            r=i/(i/l);
            ans=ans+1LL*(r-l+1)*(i/l);
        }
        sum[i]=ans;
    }
}
int main(){
    scanf("%d",&t);
    prime(50000);
    pre(50000);
    // printf("ok\n");
    while(t--){
        scanf("%d %d",&n,&m);
        long long ans=0;
        for(int l=1,r;l<=min(n,m);l=r+1){
            r=min(n/(n/l),m/(m/l));
            // printf("l=%d\n",l);
            ans=ans+1LL*(mu[r]-mu[l-1])*sum[n/l]*sum[m/l];
        }
        printf("%lld\n",ans);
    }
    return 0;
}