【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[],qy[],f[N],prime[N],cnt,mu[N],vis[N];
ll sum[N];
void calcmu(){
    cnt=;memset(vis,,sizeof(vis));mu[]=;
    for(int i=;i<=maxn;i++){
        if(vis[i]){prime[++cnt]=i;mu[i]=-;}
        for(int j=;j<=cnt;j++){
            int t=prime[j]*i;if(t>maxn)break;
            vis[t]=;
            if(i%prime[j]==){mu[t]=;break;}
            mu[t]=-mu[i];
        }
    }
    for(int i=;i<=maxn;i++)if(mu[i])sum[i]-=mu[i];
    for(int i=maxn;i>=;i--)if(sum[i])
    for(ll j=1LL*i*i;j<=maxn;j*=i)sum[j]=sum[i];
    for(int i=;i<=maxn;i++)sum[i]+=sum[i-];
}
inline int read(){
    int f=,x=;char ch;
    do{ch=getchar();if(ch=='-')f=-;}while(ch<''||ch>'');
    do{x=x*+ch-'';ch=getchar();}while(ch>=''&&ch<='');
    return f*x;
}
int main(){
    int T=read();
    for(int i=;i<=T;i++){
        qx[i]=read();qy[i]=read();
        maxn=max(maxn,max(qx[i],qy[i]));
    }
    calcmu();
    for(int aaa=;aaa<=T;aaa++){
        int n=qx[aaa],m=qy[aaa];
        if(n>m)swap(n,m);ll ans=;
        for(int i=,j=;i<=n;i=j+){
            j=min(n/(n/i),m/(m/i));
            ans+=(ll)(n/i)*(ll)(m/i)*(sum[j]-sum[i-]);
        }
        printf("%lld\n",ans);
    }
}