51nod 1222 莫比乌斯反演

思路：

yhx找的反演题

 题解已经烂大街了

#pragma GCC optimize("O3")
//By SiriusRen
#include <bits/stdc++.h>
using namespace std;
typedef long long ll;
int mu[1000500],prime[999999],vis[1000500],tot;
void init(){
    mu[1]=1;
    for(int i=2;i<=1000000;i++){
        if(!vis[i])prime[++tot]=i,mu[i]=-1;
        for(int j=1;j<=tot&&i*prime[j]<=1000000;j++){
            vis[i*prime[j]]=1;mu[i*prime[j]]=-mu[i];
            if(i%prime[j]==0){mu[i*prime[j]]=0;break;}
        }
    }
}
ll f(ll n){
    ll res=0;
    for(ll i=1;i*i*i<=n;res-=5,i++)
        for(ll j=i;i*j*j<=n;j++)
            res+=(n/i/j-j+1)*6-(i==j?(n/i/j-j)*3:3);
    return res;
}
ll solve(ll n){
    ll res=0;
    for(ll i=1;i*i<=n;i++)res+=mu[i]?mu[i]*f(n/i/i):0;
    return (res+n)/2;
}
int main(){
    init();ll a,b;
    scanf("%lld%lld",&a,&b);
    printf("%lld\n",solve(b)-solve(a-1));
}