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

$sumlimits_{i=1}^nsumlimits_{j=1}^m[gcd(i,j)==prime]$

$sum_{p∈prime}f(p)$

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

套路的，改为枚举$frac{d}{p}$

$sum_{p∈prime} sum_{d=1}^{min(leftlfloorfrac{n}{p} ight floor,leftlfloorfrac{m}{p} ight floor)}μ(d)g(dp)$

这时候可以换掉$g$了

$sum_{p∈prime} sum_{d=1}^{min(leftlfloorfrac{n}{p} ight floor,leftlfloorfrac{m}{p} ight floor)}μ(d)leftlfloorfrac{n}{dp} ight floorleftlfloorfrac{m}{dp} ight floor$

然后换回枚举现在的$dp$(即原来的$d$)，交换求和号

$sumlimits_{i=1}^{min(n,m)}sum_{d|i&&d∈prime}μ(frac{i}{d})leftlfloorfrac{n}{i} ight floorleftlfloorfrac{m}{i} ight floor$

$sumlimits_{i=1}^{min(n,m)}leftlfloorfrac{n}{i} ight floorleftlfloorfrac{m}{i} ight floorsum_{d|i&&d∈prime}μ(frac{i}{d})$

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

#include<cstdio>
#include<cstring>
#include<algorithm>
using namespace std;
const int N=10000005;
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[1]=1,npr[1]=true,maxx=10000000;
    for(i=2;i<=maxx;i++)
    {
        if(!npr[i]) pri[++cnt]=i,mul[i]=-1;
        for(j=1;j<=cnt&&1ll*i*pri[j]<=maxx;j++)
        {
            npr[i*pri[j]]=true;
            if(i%pri[j]==0) break;
            else mul[i*pri[j]]=-mul[i];
        }
    }
    for(i=1;i<=maxx;i++)
        for(j=1;j<=cnt&&1ll*i*pri[j]<=maxx;j++)
            mulsum[i*pri[j]]+=mul[i];
    for(i=1;i<=maxx;i++) mulsum[i]+=mulsum[i-1];
}
int main()
{
    register int i,j;
    scanf("%d",&T),prework();
    while(T--)
    {
        scanf("%d%d",&n,&m),ans=0,nm=min(n,m);
        for(i=1;i<=nm;i=j+1)
        {
            j=min(n/(n/i),m/(m/i));
            ans+=(mulsum[j]-mulsum[i-1])*(n/i)*(m/i);
        }
        printf("%lld
",ans);
    }
    return 0;
}