SPOJ VLATTICE Visible Lattice Points（莫比乌斯反演）题解

题意：

有一个\(n*n*n\)的三维直角坐标空间，问从\((0,0,0)\)看能看到几个点。

思路：

按题意研究一下就会发现题目所求为\(\sum_{i=0}^n\sum_{j=0}^n\sum_{k=0}^ngcd(i,j,k)==1\)，那么我们定义\(f(n)\)为\(gcd(i,j,k)==n\)的对数，\(F(n)\)为\(gcd(i,j,k)\)是\(n\)倍数的对数，显然\(F(n)\)比较容易求，那么可得\(f(1)=\sum_{1|d}\mu(\frac{d}{1})F(d)\)。

代码：

#include<map>
#include<set>
#include<queue>
#include<stack>
#include<ctime>
#include<cmath>
#include<cstdio>
#include<string>
#include<vector>
#include<cstring>
#include<sstream>
#include<iostream>
#include<algorithm>
using namespace std;
typedef long long ll;
typedef unsigned long long ull;
const int maxn = 1000000 + 5;
const int INF = 0x3f3f3f3f;
const ull seed = 131;
const ll MOD = 1e9;
using namespace std;

int mu[maxn], vis[maxn];
int prime[maxn], cnt;
void getmu(int n){
    memset(vis, 0, sizeof(vis));
    memset(mu, 0, sizeof(mu));
    cnt = 0;
    mu[1] = 1;
    for(int i = 2; i <= n; i++) {
        if(!vis[i]){
            prime[cnt++] = i;
            mu[i] = -1;
        }
        for(int j = 0; j < cnt && prime[j] * i <= n; j++){
            vis[prime[j] * i] = 1;
            if(i % prime[j] == 0) break;
            mu[i * prime[j]] = -mu[i];
        }
    }
}
ll get(int n){
    return 1LL * n * n * n + 3LL * n * n + 3LL * n;
}
int main(){
    int n, T;
    getmu(1000000);
    scanf("%d", &T);
    while(T--){
        scanf("%d", &n);
        ll ans = 0;
        for(int i = 1; i <= n; i++){
            ans += 1LL * mu[i] * get(n / i);
        }
        printf("%lld\n", ans);
    }
    return 0;
}