【反演复习计划】【bzoj3994】DZY loves maths

这题大概就是提取一下d，然后就跟前面的题目差不多了。

#include<bits/stdc++.h>
#define N 10000005
using namespace std;
typedef long long ll;
int maxn,qx[100010],qy[100010],f[N],prime[N],cnt,mu[N],vis[N];
ll sum[N];
void calcmu(){
    cnt=0;memset(vis,1,sizeof(vis));mu[1]=1;
    for(int i=2;i<=maxn;i++){
        if(vis[i]){prime[++cnt]=i;mu[i]=-1;}
        for(int j=1;j<=cnt;j++){
            int t=prime[j]*i;if(t>maxn)break;
            vis[t]=0;
            if(i%prime[j]==0){mu[t]=0;break;}
            mu[t]=-mu[i];
        }
    }
    for(int i=2;i<=maxn;i++)if(mu[i])sum[i]-=mu[i];
    for(int i=maxn;i>=2;i--)if(sum[i])
    for(ll j=1LL*i*i;j<=maxn;j*=i)sum[j]=sum[i];
    for(int i=2;i<=maxn;i++)sum[i]+=sum[i-1];
}
inline int read(){
    int f=1,x=0;char ch;
    do{ch=getchar();if(ch=='-')f=-1;}while(ch<'0'||ch>'9');
    do{x=x*10+ch-'0';ch=getchar();}while(ch>='0'&&ch<='9');
    return f*x;
}
int main(){
    int T=read();
    for(int i=1;i<=T;i++){
        qx[i]=read();qy[i]=read();
        maxn=max(maxn,max(qx[i],qy[i]));
    }
    calcmu();
    for(int aaa=1;aaa<=T;aaa++){
        int n=qx[aaa],m=qy[aaa];
        if(n>m)swap(n,m);ll ans=0;
        for(int i=1,j=1;i<=n;i=j+1){
            j=min(n/(n/i),m/(m/i));
            ans+=(ll)(n/i)*(ll)(m/i)*(sum[j]-sum[i-1]);
        }
        printf("%lld\n",ans);
    }
}