BZOJ 3994: [SDOI2015]约数个数和3994: [SDOI2015]约数个数和 莫比乌斯反演

https://www.lydsy.com/JudgeOnline/problem.php?id=3994

https://blog.csdn.net/qq_36808030/article/details/77056706

莫比乌斯反演，我现在莫比乌斯反演都不会写不会推了。

#include<iostream>
#include<cstdio>
#include<cstring>
#include<algorithm>
#include<cmath>
#include<queue>
using namespace std;
#define LL long long
const int maxn=50010;
int mu[maxn]={},pri[maxn]={},tot=0;
bool v[maxn]={};LL f[maxn]={};
void get_mu(int n){
    mu[1]=1;
    for(int i=2;i<=n;++i){
        if(!v[i]){pri[++tot]=i;mu[i]=-1;}
        for(int j=1;j<=tot&&pri[j]*i<=n;++j){
            int w=pri[j]*i;v[w]=1;
            if(i%pri[j]==0){mu[w]=0;break;}
            mu[w]=mu[i]*(-1);
        }
    }
    for(int i=1;i<=n;++i){
        mu[i]+=mu[i-1];
        for(int j=1,las;j<=i;j=las+1){
            las=i/(i/j);
            f[i]+=(LL)(las-j+1)*(LL)(i/j);
        }
    }
}
int main(){
    get_mu(maxn-10);
    int T;scanf("%d\n",&T);
    while(T-->0){
        int n,m; scanf("%d%d",&n,&m);
        if(n>m)swap(n,m);
        LL ans=0;
        for(int i=1,las;i<=n;i=las+1){
            las=min(n/(n/i),m/(m/i));
            ans+=(LL)(mu[las]-mu[i-1])*f[n/i]*f[m/i];
        }
        printf("%lld\n",ans);
    }
    return 0;
}