SPOJ 7001. Visible Lattice Points （莫比乌斯反演）

7001. Visible Lattice Points

Problem code: VLATTICE

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

这题就是求gcd(a,b,c) = 1    a,b,c <=N 的对数。

用莫比乌斯反演可以求解。

设g(n)为gcd(x,y,z)=n的个数，f(n)为n | g(i*n)的个数，那么有f(n)=sigma(n|d,g(d))，那么g(n)=sigma(n|d, mu(d/n)*f(d))，我们要求g(1)，则g(1)=sigma(n|d, mu(d)*f(d))，

因为f(d)=(n/d)*(n/d)*(n/d)，所以g(1)=sigma( mu(d)*(n/d)*(n/d)*(n/d) ).

 /* ***********************************************
 Author        :kuangbin
 Created Time  :2013/8/21 18:28:50
 File Name     :F:\2013ACM练习\专题学习\数学\莫比乌斯反演\SPOJ7001.cpp
 ************************************************ */

 #include <stdio.h>
 #include <string.h>
 #include <iostream>
 #include <algorithm>
 #include <vector>
 #include <queue>
 #include <set>
 #include <map>
 #include <string>
 #include <math.h>
 #include <stdlib.h>
 #include <time.h>
 using namespace std;
 const int MAXN = ;
 bool check[MAXN+];
 int prime[MAXN+];
 int mu[MAXN+];
 void Moblus()
 {
     memset(check,false,sizeof(check));
     mu[] = ;
     int tot = ;
     for(int i = ; i <= MAXN; i++)
     {
         if( !check[i] )
         {
             prime[tot++] = i;
             mu[i] = -;
         }
         for(int j = ; j < tot; j++)
         {
             if(i * prime[j] > MAXN) break;
             check[i * prime[j]] = true;
             if( i % prime[j] == )
             {
                 mu[i * prime[j]] = ;
                 break;
             }
             else
             {
                 mu[i * prime[j]] = -mu[i];
             }
         }
     }
 }
 int main()
 {
     //freopen("in.txt","r",stdin);
     //freopen("out.txt","w",stdout);
     int T,n;
     Moblus();
     scanf("%d",&T);
     while(T--)
     {
         scanf("%d",&n);
         long long ans = ;
         for(int i = ;i <= n;i++)
             ans += (long long)mu[i]*(n/i)*(n/i)*((n/i)+);
         printf("%lld\n",ans);
     }
     return ;
 }