P3327/bzoj3994 [SDOI2015]约数个数和（莫比乌斯反演）

P3327 [SDOI2015]约数个数和

神犇题解（转）

无话可补

#include<iostream>
#include<cstdio>
#include<cstring>
#define re register
using namespace std;
template<typename T>T max(T &a,T &b){return a>b?a:b;}
template<typename T>T min(T &a,T &b){return a<b?a:b;}
#define N 50001
int t,n,m,pct,pri[N],mu[N],sum[N];
long long g[N],ans;
bool v[N];
int main(){
    mu[]=;
    for(re int i=;i<N;++i){
        if(!v[i]) pri[++pct]=i,mu[i]=-;
        for(re int j=;j<=pct;++j){
            re int tmp=i*pri[j];
            if(tmp>=N) break;
            v[tmp]=;
            if(i%pri[j]) mu[tmp]=-mu[i];
            else break;
        }//线性筛
    }re int u;
    for(u=;u+<N;u+=){
        sum[u]=sum[u-]+mu[u];
        sum[u+]=sum[u]+mu[u+];
        sum[u+]=sum[u+]+mu[u+];
        sum[u+]=sum[u+]+mu[u+];
    }//循环展开：微小加速
    for(;u<N;++u) sum[u]=sum[u-]+mu[u];
    for(re int i=;i<N;++i){
        ans=;
        for(re int l=,r;l<=i;l=r+){
            r=i/(i/l);
            ans+=1ll*(r-l+)*(i/l);
        }g[i]=ans;
    }

    scanf("%d",&t);
    while(t--){
        scanf("%d%d",&n,&m);
        if(n>m) swap(n,m);
        ans=;
        for(re int l=,r;l<=n;l=r+){
            r=min(n/(n/l),m/(m/l));
            ans+=1ll*(sum[r]-sum[l-])*g[n/l]*g[m/l];
        }printf("%lld\n",ans);
    }return ;
}