51nod 1222 最小公倍数计数【莫比乌斯反演】

参考：https://www.geek-share.com/detail/2708787922.html
所是反演其实反演作用不大，又是一道做起来感觉诡异的题
转成前缀和相减的形式
\[ \sum_{i=1}^{n}\sum_{j=1}^{n}[\frac{i*j}{gcd(i,j)}\leq n] \]
\[ \sum_{d=1}^{n}\sum_{i=1}^{\left \lfloor \frac{n}{d}\right \rfloor}\sum_{j=1}^{\left \lfloor \frac{n}{d}\right \rfloor}[gcd(i,j)==1][i*j\leq\left \lfloor \frac{n}{d} \right \rfloor] \]
\[ \sum_{k=1}^{n} \mu(k)\sum_{d=1}^{\left \lfloor \frac{n}{k} \right \rfloor}\sum_{i=1}^{\left \lfloor \frac{n}{dk} \right \rfloor}\sum_{j=1}^{\left \lfloor \frac{n}{dk} \right \rfloor}[i*j*d\leq\left \lfloor \frac{n}{k^2} \right \rfloor] \]
然后是非常神奇的缩小范围……
\[ \sum_{k=1}^{\sqrt{n}}\mu(k)\sum_{d=1}^{\left \lfloor \frac{n}{k^2} \right \rfloor}\sum_{i=1}^{\left \lfloor \frac{n}{dk^2} \right \rfloor}\sum_{j=1}^{\left \lfloor \frac{n}{dk^2} \right \rfloor}[i*j*d\leq\left \lfloor \frac{n}{k^2} \right \rfloor] \]
然后对于这个友好的范围直接枚举就可以了。

#include<iostream>
#include<cstdio>
#include<cmath>
using namespace std;
const int N=1000005,m=1000000;
int q
,mb
,tot;
long long a,b;
bool v
;
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;
}