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

题意：GCD(a,b,c)=1,   0<=a,b,c<=N ;

莫比乌斯反演，十分的巧妙。

GCD(a,b)的题十分经典。这题扩展到GCD(a,b,c)加了一维，但是思想却是相同的。

设f(d) = GCD(a,b,c) = d的种类数 ；

F(n) 为GCD(a,b,c) = d 的倍数的种类数, n%a == 0 n%b==0 n%c==0。

即 ：F(d) = (N/d)*(N/d)*(N/d);

则f(d) = sigma( mu[n/d]*F(n), d|n )

由于d = 1 所以f(1) = sigma( mu[n]*F(n) ) = sigma( mu[n]*(N/n)*(N/n)*(N/n) );

由于0能够取到，所以对于a,b,c 要讨论一个为0 ，两个为0的情况 (3种).

 #include<iostream>
 #include<stdio.h>
 #include<cstring>
 #include<cstdlib>
 using namespace std; typedef long long LL;
 const int maxn = +;
 bool s[maxn];
 int prime[maxn],len = ;
 int mu[maxn];
 void  init()
 {
     memset(s,true,sizeof(s));
     mu[] = ;
     for(int i=;i<maxn;i++)
     {
         if(s[i] == true)
         {
             prime[++len]  = i;
             mu[i] = -;
         }
         for(int j=;j<=len && (long long)prime[j]*i<maxn;j++)
         {
             s[i*prime[j]] = false;
             if(i%prime[j]!=)
                 mu[i*prime[j]] = -mu[i];
             else
             {
                 mu[i*prime[j]] = ;
                 break;
             }
         }
     }
 } int main()
 {
     int n,T;
     init();
     scanf("%d",&T);
     while(T--)
     {
         scanf("%d",&n);
         LL sum = ;
         for(int i=;i<=n;i++)
             sum = sum + (long long)mu[i]*(n/i)*(n/i)*;
         for(int i=;i<=n;i++)
             sum = sum + (long long)mu[i]*(n/i)*(n/i)*(n/i);
         printf("%lld\n",sum);
     }
     return ;
 }