51nod 1222 最小公倍数计数【莫比乌斯反演】

参考：https://www.cnblogs.com/SilverNebula/p/7045199.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[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;
}