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).

#include<iostream>
#include<stdio.h>
#include<cstring>
#include<cstdlib>
using namespace std;

typedef long long LL;
const int maxn = 1000000+3;
bool s[maxn];
int prime[maxn],len = 0;
int mu[maxn];
void  init()
{
    memset(s,true,sizeof(s));
    mu[1] = 1;
    for(int i=2;i<maxn;i++)
    {
        if(s[i] == true)
        {
            prime[++len]  = i;
            mu[i] = -1;
        }
        for(int j=1;j<=len && (long long)prime[j]*i<maxn;j++)
        {
            s[i*prime[j]] = false;
            if(i%prime[j]!=0)
                mu[i*prime[j]] = -mu[i];
            else
            {
                mu[i*prime[j]] = 0;
                break;
            }
        }
    }
}

int main()
{
    int n,T;
    init();
    scanf("%d",&T);
    while(T--)
    {
        scanf("%d",&n);
        LL sum = 3;
        for(int i=1;i<=n;i++)
            sum = sum + (long long)mu[i]*(n/i)*(n/i)*3;
        for(int i=1;i<=n;i++)
            sum = sum + (long long)mu[i]*(n/i)*(n/i)*(n/i);
        printf("%lld
",sum);
    }
    return 0;
}