Visible Lattice Points（莫比乌斯反演）

题目：

Consider a N*N*N lattice. One corner is at (0,0,0) and the opposite one is at (N,N,N). How many lattice points are visible from corner at (0,0,0) ? A point X is visible from point Y iff no other lattice point lies on the segment joining X and Y. 
 
Input : 
The first line contains the number of test cases T. The next T lines contain an interger N 
 
Output : 
Output T lines, one corresponding to each test case. 
 
Sample Input : 
3 
1 
2 
5 
 
Sample Output : 
7 
19 
175 
 
Constraints : 
T <= 50 
1 <= N <= 1000000

解题报告：

关于莫比乌斯反演求这个gcd的问题，重要的就是理解怎么来的，这就稍微有点难度了，代码不是很难实现，关于这道题目我参考的这位大佬的博客：https://blog.csdn.net/baiyifeifei/article/details/82954002，推导过程比较详细了，

我觉得就是要掌握一个全局和部分的关系，才可以实现的。

ac代码：

//from::Onion
//kuangbin Visible Lattice Points 莫比乌斯反演 
#include<iostream>
#include<algorithm>
#include<cstring>
#include<cstdio>
#include<cmath>
using namespace std;
typedef long long ll;

const int maxn =1e6+1000;
int prime[maxn];
int mu[maxn];
bool vis[maxn];
int cnt;
ll n;

void db_Onion()
{
    cnt=0;
    mu[1]=1;
    vis[0]=vis[1]=1;
    for(int i=2;i<maxn;i++)
    {
        if(!vis[i])
        {
            prime[cnt++]=i;
            mu[i]=-1;
        }
        for(int j=0;j<cnt&&prime[j]*i<maxn;j++)
        {
            vis[prime[j]*i]=1;
            if(i%prime[j]==0)
            {
                mu[i*prime[j]]=0;
            }
            else
            {
                mu[i*prime[j]]=-mu[i];
            }
        }
    }
}
int main()
{
    db_Onion();
    int t;
    scanf("%d",&t);
    while(t--)
    {
        scanf("%d",&n);
        ll ans=0;
        for(int i=1;i<=n;i++)
        {
            ll tmp=floor(n/i);
            ans+=(mu[i]*tmp*tmp*tmp)+3*(mu[i]*tmp*tmp);
        }    
        printf("%lld
",ans+3);        
    } 
}