SPOJ VLATTICE (莫比乌斯反演)

传送门:https://www.spoj.com/problems/VLATTICE/en/

题意:

在三维坐标系下,你在点(0,0,0),看的范围是(n,n,n)以内,求你可以看见多少个点没有被遮挡

题解:

一条线上的点肯定是会被挡的

所以我们求的是\(gcd(x,y,z)==1\)的组数

我们设

\[f(d):gcd(x,y,z)=d的对数\\ F(d):d|gcd(x,y,z)的对数\\ 由于F(d)为[n/d]*[n/d]*[n/d]\\ 所以反演可得\\ f(1)=mu[d]*[n/d]*[n/d]*[n/d]\\ 由于坐标系上的点也要算的话\\ 1.我们的点(0,0,1)、(1,0,1)、(0,1,1)\\ 2.xoy,xoz,xoy面上的点gcd(i,j)==1; \\ 3.其他点 gcd(i,j,k)==1 \]

代码:

#include <set>
#include <map>
#include <deque>
#include <queue>
#include <stack>
#include <cmath>
#include <ctime>
#include <bitset>
#include <cstdio>
#include <string>
#include <vector>
#include <cstdlib>
#include <cstring>
#include <iostream>
#include <algorithm>
using namespace std;

typedef long long LL;
typedef long long ll;
typedef pair<LL, LL> pLL;
typedef pair<LL, int> pLi;
typedef pair<int, LL> pil;;
typedef pair<int, int> pii;
typedef unsigned long long uLL;
#define ls rt<<1
#define rs rt<<1|1
#define lson l,mid,rt<<1
#define rson mid+1,r,rt<<1|1
#define bug printf("*********\n")
#define FIN freopen("input.txt","r",stdin);
#define FON freopen("output.txt","w+",stdout);
#define IO ios::sync_with_stdio(false),cin.tie(0)
#define debug1(x) cout<<"["<<#x<<" "<<(x)<<"]\n"
#define debug2(x,y) cout<<"["<<#x<<" "<<(x)<<" "<<#y<<" "<<(y)<<"]\n"
#define debug3(x,y,z) cout<<"["<<#x<<" "<<(x)<<" "<<#y<<" "<<(y)<<" "<<#z<<" "<<z<<"]\n"

const double eps = 1e-8;
const int mod = 1e9 + 7;
const int maxn = 1e6 + 5;
const int INF = 0x3f3f3f3f;
const LL INFLL = 0x3f3f3f3f3f3f3f3f;

int mu[maxn];
int prime[maxn];
int not_prime[maxn];
int tot;
void Mobiwus(int n) {
    mu[1] = 1;
    for(int i = 2; i <= n; i++) {
        if(!not_prime[i]) {
            prime[++tot] = i;
            mu[i] = -1;
        }
        for(int j = 1; prime[j]*i <= n; j++) {
            not_prime[prime[j]*i] = 1;
            if(i % prime[j] == 0) {
                mu[prime[j]*i] = 0;
                break;
            }
            mu[prime[j]*i] = -mu[i];
        }
    }
}
int main() {
#ifndef ONLINE_JUDGE
    FIN
#endif
    int T;
    Mobiwus(1000005);
    scanf("%d", &T);
    while(T--) {
        int n;
        scanf("%d", &n);
        LL ans = 3;
        for(int i = 1; i <= n; i++) {
            ans += 1LL * mu[i] * (n / i) * (n / i) * (n / i);
        }
        for(int i = 1; i <= n; i++) {
            ans += 1LL * mu[i] * (n / i) * (n / i) * 3;
        }
        printf("%lld\n", ans);
    }
    return 0;
}
posted @ 2019-06-10 11:04  buerdepepeqi  阅读(163)  评论(0编辑  收藏  举报