3994: [SDOI2015]约数个数和

#include<cstdio>
#include<iostream>
using namespace std;
typedef long long ll;
const int F=1005;
const int N=5e4+5;
int n,m,T,tot,prime[N/3],mu[N],c[N];bool check[N];
ll ANS[F][F],sum[N],f[N];
inline int read(){
    int x=0,f=1;char ch=getchar();
    while(ch<‘0‘||ch>‘9‘){if(ch==‘-‘)f=-1;ch=getchar();}
    while(ch>=‘0‘&&ch<=‘9‘){x=x*10+ch-‘0‘;ch=getchar();}
    return x*f;
}
void prepare(int n){
    mu[1]=1;f[1]=1;c[1]=1;
    for(int i=2;i<=n;i++){
        if(!check[i]){
            prime[++tot]=i;
            f[i]=2;
            c[i]=1;
            mu[i]=-1;
        } 
        for(int j=1;j<=tot&&i*prime[j]<=n;j++){
            check[i*prime[j]]=1;
            if(!(i%prime[j])){
                f[i*prime[j]]=f[i]/(c[i]+1)*(c[i]+2);
                c[i*prime[j]]=c[i]+1;
                mu[i*prime[j]]=0;
                break;
            }
            else{
                f[i*prime[j]]=f[i]*f[prime[j]];
                c[i*prime[j]]=1;
                mu[i*prime[j]]=-mu[i];
            } 
        }
    }
    for(int i=1;i<=n;i++) sum[i]=sum[i-1]+mu[i];
    for(int i=1;i<=n;i++) f[i]+=f[i-1];
}
void solve(){
    if(n<=1000&&m<=1000) if(ANS[n][m]){
        printf("%lld\n",ANS[n][m]);
        return ;
    }
    ll ans(0);
    for(int i=1,pos;i<=n;i=pos+1){
        pos=min(n/(n/i),m/(m/i));
        ans+=(sum[pos]-sum[i-1])*f[n/i]*f[m/i];
    }
    if(n<=1000&&m<=1000) ANS[n][m]=ANS[m][n]=ans;
    printf("%lld\n",ans);
}
int main(){
    prepare(50000);
    T=read();
    while(T--){
        n=read();m=read();
        if(n>m) swap(n,m);
        solve();
    }
    return 0;
}