spoj 7001. Visible Lattice Points GCD问题 莫比乌斯反演

SPOJ Problem Set (classical) 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

 

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).

 1 #include<iostream>

 2 #include<stdio.h>

 3 #include<cstring>

 4 #include<cstdlib>

 5 using namespace std;

 6 

 7 typedef long long LL;

 8 const int maxn = 1000000+3;

 9 bool s[maxn];

10 int prime[maxn],len = 0;

11 int mu[maxn];

12 void  init()

13 {

14     memset(s,true,sizeof(s));

15     mu[1] = 1;

16     for(int i=2;i<maxn;i++)

17     {

18         if(s[i] == true)

19         {

20             prime[++len]  = i;

21             mu[i] = -1;

22         }

23         for(int j=1;j<=len && (long long)prime[j]*i<maxn;j++)

24         {

25             s[i*prime[j]] = false;

26             if(i%prime[j]!=0)

27                 mu[i*prime[j]] = -mu[i];

28             else

29             {

30                 mu[i*prime[j]] = 0;

31                 break;

32             }

33         }

34     }

35 }

36 

37 int main()

38 {

39     int n,T;

40     init();

41     scanf("%d",&T);

42     while(T--)

43     {

44         scanf("%d",&n);

45         LL sum = 3;

46         for(int i=1;i<=n;i++)

47             sum = sum + (long long)mu[i]*(n/i)*(n/i)*3;

48         for(int i=1;i<=n;i++)

49             sum = sum + (long long)mu[i]*(n/i)*(n/i)*(n/i);

50         printf("%lld\n",sum);

51     }

52     return 0;

53 }