參考：https://www.cnblogs.com/SilverNebula/p/7045199.html
所是反演其實反演作用不大，又是一道做起來感覺詭異的題
轉成前綴和相減的形式
\[ \sum_{i=1}^{n}\sum_{j=1}^{n}[\frac{i*j}{gcd(i,j)}\leq n] \]
\[ \sum_{d=1}^{n}\sum_{i=1}^{\left \lfloor \frac{n}{d}\right \rfloor}\sum_{j=1}^{\left \lfloor \frac{n}{d}\right \rfloor}[gcd(i,j)==1][i*j\leq\left \lfloor \frac{n}{d} \right \rfloor] \]

#include<iostream>
#include<cstdio>
#include<cmath>
using namespace std;
const int N=1000005,m=1000000;
int q[N],mb[N],tot;
long long a,b;
bool v[N];
long long wk(long long n)
{
    if(!n)
        return 0;
    long long re=0ll,tmp=0ll,a=sqrt(n);
    for(long long k=1;k<=a;k++)
        if
 
(mb[k])
        {
            tmp=0;
            long long b=n/k/k;
            for(long long i=1;i*i*i<=b;i++)
            {
                for(long long j=i+1;j*j*i<=b;j++)
                    tmp+=(b/(i*j)-j)*6+3;
                tmp+=(b/(i*i)-i)*3;
                tmp++;
            }
            re+=mb[k]*tmp;
        }
    return (re+n)/2;
}       
int main()
{
    mb[1]=1;
    for(int i=2;i<=m;i++)
    {
        if(!v[i])
        {
            q[++tot]=i;
            mb[i]=-1;
        }
        for(int j=1;j<=tot&&i*q[j]<=m;j++)
        {
            int k=i*q[j];
            v[k]=1;
            if(i%q[j]==0)
            {
                mb[k]=0;
                break;
            }
            mb[k]=-mb[i];
        }
    }
    scanf("%lld%lld",&a,&b);
    printf("%lld\n",wk(b)-wk(a-1));
    return 0;
}

51nod 1222 最小公倍數計數【莫比烏斯反演】