SPOJ 7001 Visible Lattice Points (数论关于gcd,超经典极力推荐-莫比乌斯反演)

传送门:http://www.spoj.com/problems/VLATTICE/

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 : 




 
Sample Output : 

19 
175 
 
Constraints : 
T <= 50 
1 <= N <= 1000000

Added by: Varun Jalan
Date: 2010-07-29
Time limit: 7s
Source limit: 50000B
Memory limit: 256MB
Cluster: Pyramid (Intel Pentium III 733 MHz)
Languages: All except: NODEJS PERL 6
Resource: own problem used for Indian ICPC training camp








题意:0<=x,y,z<=n,求有多少对xyz满足gcd(x,y,z)=1。

题解:需要用到莫比乌斯反演,好久之前都看过这定律,直到今天做了这道题才弄懂。


首先我们介绍下莫比乌斯反演:

g(n)=sigma(d|n,f(d))

f(n)=sigma(d|n,u(d)*g(n/d))

u(d)定义

若d=1    那么μ(d)=1
若d=p1p2…pr    (r个不同质数,且次数都唯一)μ(d)=(-1)^r
其余    μ(d)=0

其实还有一种写法:

g(n)=sigma(d|n,f(d))

f(n)=sigma(n|d,u(d/n)*g(d))


这里我们设g(x)为满足x | (i, j, k)的个数,f(x)为满足(i, j, k) = x的个数。

所以g(n)=sigma(d|n, f(d)) = [n/x] * [n/x] * [n/x]。

而题目要求f(1)等于多少,f(1)=sigma(1|d,u(d)*g(d))。


所以现在关键的是求u(d),可以通过线性筛选素数的方法求解u,详细请看下面代码。


然后还要加上三个平面上面的gcd,直接mu[i]*(n/i)*(n/i)*(n/i+3),二维的比三维的少一维-三个平面。

别忘了还有x,y,z三个轴。


好了,到这里全部的算法都以说清楚了……


AC代码:

10188459 2013-10-06 09:31:55 xiaohao Visible Lattice Points accepted
edit  run
4.65 15M

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

#define si1(a) scanf("%d",&a)
#define si2(a,b) scanf("%d%d",&a,&b)
#define sd1(a) scanf("%lf",&a)
#define sd2(a,b) scanf("%lf%lf",&a,&b)
#define ss1(s)  scanf("%s",s)
#define pi1(a)    printf("%d\n",a)
#define pi2(a,b)  printf("%d %d\n",a,b)
#define mset(a,b)   memset(a,b,sizeof(a))
#define forb(i,a,b)   for(int i=a;i<b;i++)
#define ford(i,a,b)   for(int i=a;i<=b;i++)

typedef long long LL;
const int N=1000010;
const int INF=0x3f3f3f3f;
const double PI=acos(-1.0);
const double eps=1e-7;


int mu[N];
LL x,pri[N];
bool vis[N];

void xiaohao_mu()//求莫比乌斯反演中的mu
{
    mset(vis,0);mset(mu,0);
    x=0;
    LL n=1000000;
    mu[1]=1;
    for(LL i=2;i<=n;i++)
    {
        if(!vis[i])//筛选素数法
        {
            pri[x++]=i;
            mu[i]=-1;//素数一定等于-1
        }
        for(LL j=0;j<x&&i*pri[j]<=n;j++)
        {
            vis[i*pri[j]]=1;
            if(i%pri[j])    mu[i*pri[j]]=-mu[i];
            else
            {
                mu[i*pri[j]]=0;
                break;
            }
        }
    }
}


int main()
{
    xiaohao_mu();
    int T;
    LL n;
    cin>>T;
    while(T--)
    {
        cin>>n;
        LL sum=3;//x,y,z轴上面的3个
        for(LL i=1;i<=n;i++)
            sum+=mu[i]*(n/i)*(n/i)*(n/i+3);
        cout<<sum<<endl;
    }
    return 0;
}


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