SPOJ7001(SummerTrainingDay04-N 莫比乌斯反演)

Visible Lattice Points

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

 //2017-08-04
 #include <cstdio>
 #include <cstring>
 #include <iostream>
 #include <algorithm>

 using namespace std;

 const int N = ;
 int mu[N], prime[N], tot, phi[N];
 long long plane[N];
 bool book[N];

 void Moblus()//求出莫比乌斯函数
 {
     memset(book,false,sizeof(book));
     mu[] = ;
     int tot = ;
     for(int i = ; i <= N; i++){
         if(!book[i]){
             prime[tot++] = i;
             mu[i] = -;
         }
         for(int j = ; j < tot; j++){
             if(i * prime[j] > N) break;
             book[i * prime[j]] = true;
             if( i % prime[j] == ){
                 mu[i * prime[j]] = ;
                 break;
             }else{
                 mu[i * prime[j]] = -mu[i];
             }
         }
     }
 }

 void getphi()
 {
    int i,j;
    phi[]=;
    for(i=;i<=N;i++)//相当于分解质因式的逆过程
    {
        if(!book[i])
        {
             prime[++tot]=i;//筛素数的时候首先会判断i是否是素数。
             phi[i]=i-;//当 i 是素数时 phi[i]=i-1
         }
        for(j=;j<=tot;j++)
        {
           if(i*prime[j]>N)  break;
           book[i*prime[j]]=;//确定i*prime[j]不是素数
           if(i%prime[j]==)//接着我们会看prime[j]是否是i的约数
           {
              phi[i*prime[j]]=phi[i]*prime[j];break;
           }
           else  phi[i*prime[j]]=phi[i]*(prime[j]-);//其实这里prime[j]-1就是phi[prime[j]]，利用了欧拉函数的积性
        }
    }
 }

 int main()
 {
     int T, n;
     scanf("%d", &T);
     Moblus();
     getphi();
     plane[] = ;
     for(int i = ; i < N; i++){
         plane[i] = plane[i-]+phi[i];
     }
     while(T--){
         scanf("%d", &n);
         long long ans = ;
         for(int d = ; d <= n; d++){
             int tmp = (int)(n/d);
             ans += (long long)mu[d]*tmp*tmp*tmp;
         }
         ans += *(plane[n]*+);
         printf("%lld\n", ans);
     }

     return ;
 }