解题：洛谷2257 YY的GCD

题面

初见莫比乌斯反演

有一个套路是关于GCD的反演经常设$f(d)=\sum_{gcd(i,j)==d},g(d)=\sum_{d|gcd(i,j)}$，然后推推推

$\sum\limits_{i=1}^n\sum\limits_{j=1}^m[gcd(i,j)==prime]$

$\sum_{p∈prime}f(p)$

$\sum_{p∈prime} \sum_{d=1}^{min(n,m)} [p|d] μ(\frac{d}{p})g(d)$

套路的，改为枚举$\frac{d}{p}$

$\sum_{p∈prime} \sum_{d=1}^{min(\left\lfloor\frac{n}{p}\right\rfloor,\left\lfloor\frac{m}{p}\right\rfloor)}μ(d)g(dp)$

这时候可以换掉$g$了

$\sum_{p∈prime} \sum_{d=1}^{min(\left\lfloor\frac{n}{p}\right\rfloor,\left\lfloor\frac{m}{p}\right\rfloor)}μ(d)\left\lfloor\frac{n}{dp}\right\rfloor\left\lfloor\frac{m}{dp}\right\rfloor$

然后换回枚举现在的$dp$(即原来的$d$)，交换求和号

$\sum\limits_{i=1}^{min(n,m)}\sum_{d|i\&\&d∈prime}μ(\frac{i}{d})\left\lfloor\frac{n}{i}\right\rfloor\left\lfloor\frac{m}{i}\right\rfloor$

$\sum\limits_{i=1}^{min(n,m)}\left\lfloor\frac{n}{i}\right\rfloor\left\lfloor\frac{m}{i}\right\rfloor\sum_{d|i\&\&d∈prime}μ(\frac{i}{d})$

于是数论分块，前面的直接算，后面的线性筛之后$O(n\log n)$预处理一下再做个前缀和

 #include<cstdio>
 #include<cstring>
 #include<algorithm>
 using namespace std;
 const int N=;
 int npr[N],pri[N],mul[N];
 long long mulsum[N],ans;
 int T,n,m,nm,cnt,maxx;
 void prework()
 {
     register int i,j;
     mul[]=,npr[]=true,maxx=;
     for(i=;i<=maxx;i++)
     {
         if(!npr[i]) pri[++cnt]=i,mul[i]=-;
         for(j=;j<=cnt&&1ll*i*pri[j]<=maxx;j++)
         {
             npr[i*pri[j]]=true;
             if(i%pri[j]==) break;
             else mul[i*pri[j]]=-mul[i];
         }
     }
     for(i=;i<=maxx;i++)
         for(j=;j<=cnt&&1ll*i*pri[j]<=maxx;j++)
             mulsum[i*pri[j]]+=mul[i];
     for(i=;i<=maxx;i++) mulsum[i]+=mulsum[i-];
 }
 int main()
 {
     register int i,j;
     scanf("%d",&T),prework();
     while(T--)
     {
         scanf("%d%d",&n,&m),ans=,nm=min(n,m);
         for(i=;i<=nm;i=j+)
         {
             j=min(n/(n/i),m/(m/i));
             ans+=(mulsum[j]-mulsum[i-])*(n/i)*(m/i);
         }
         printf("%lld\n",ans);
     }
     return ;
 }