SPOJ - VLATTICE (莫比乌斯反演)

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

 

题意就是给你一个三维的地图，坐标为(0,0,0)∼(n,n,n),判断有多少个坐标与原点之间的连线不经过其他的点。 

思路:统计答案的点分为三类

1.坐标轴上的点(1,0,0)(0,1,0)(0,0,1) 三个

2.xoy,xoz,xoy面上的点gcd(i,j)==1; 二维很简单

3.其他点 gcd(i,j,k)==1

 

    

代码如下:

#include <bits/stdc++.h>

using namespace std;
typedef long long ll;
const int maxn = 1000000+50;
int p[maxn],mo[maxn],phi[maxn],cnt=0;
bool vis[maxn];
void init()
{
    mo[1]=1;
    phi[1]=1;
    for(int i=2;i<=maxn-10;i++){
        if(!vis[i]){
            mo[i]=-1;
            phi[i]=i-1;
            p[cnt++]=i;
        }
        for(int j=0;j<cnt&&(ll)i*p[j]<=maxn-10;j++){
            vis[i*p[j]]=true;
            if(i%p[j]==0){
                mo[i*p[j]]=0;
                phi[i*p[j]]=phi[i]*p[j];
                break;
            }
            mo[i*p[j]]=-mo[i];
            phi[i*p[j]]=phi[i]*(p[j]-1);
        }
    }
}
int n;
int main()
{
    //freopen("de.txt","r",stdin);
    init();
    int T;
    scanf("%d",&T);
    while (T--){
        scanf("%d",&n);
        ll ans = 0;
        for (int i=1;i<=n;++i){
            ans+=(ll)mo[i]*(n/i)*(n/i)*(n/i);
        }
        for (int i=1;i<=n;++i){
            ans+=(ll)mo[i]*(n/i)*(n/i)*3;
        }
        printf("%lld\n",ans+3);
    }
    return 0;
}