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

题目大意

给定n*n*n的立方体，每一个整数点除（0。0。0）之外都有一盏灯（抽象理解），问能看到多少盏灯（被盖住的灯不算）

解题思路

莫比乌斯反演/容斥原理的典型应用

用容斥原理来解释就是三个点都能被k整除的个数乘上莫比乌斯系数，求和就可以

而三个点都能被k整除的个数就是floor(n/i)^3

注意到最大数据量为1000000 直接线性处理的办法可能TLE

而(n/i)在后面i>(n/2)的部分结果都为1 能够省去一次次计算，直接按mu的前缀和来处理

则我们就统计同样(n/i)的值是否出现两次。假设出现两次那么我们就開始依照前缀和的方法来处理

不优化 6200ms

优化后 490ms 

code

优化前

#include <cstdio>
#include <iostream>
#include <algorithm>
#include <ctime>
#include <cctype>
#include <cmath>
#include <string>
#include <cstring>
#include <stack>
#include <queue>
#include <list>
#include <vector>
#include <map>
#include <set>

#define sqr(x) ((x)*(x))
#define LL long long 
#define INF 0x3f3f3f3f
#define PI acos(-1.0)
#define eps 1e-10
#define mod 100000007ll
using namespace std;
LL n;
LL com[1000005],pri[1000005],phi[1000005],pn,sum[1000005],mu[1000005];
LL a[1000005];
int main()
{
    memset(com ,0 ,sizeof com);
    mu[1]=1;
    for (int i=2;i<=1000000ll;i++)
    {
        if (com[i]==0)
        {
            phi[i]=i-1;
            pri[++pn]=i;
            mu[i]=-1;
            // printf("%d\n", pri[pn]);
            // system("pause");
        }
        for (int j=1;j<=pn&&pri[j]*i<=1000000ll;j++)
        {
            if (i%pri[j])
            {
                phi[i*pri[j]]=phi[i]*(pri[j]-1);
                com[i*pri[j]]=1;
                mu[i*pri[j]]=-mu[i];
            }
            else 
            {
                phi[i*pri[j]]=phi[i]*(pri[j]);
                com[i*pri[j]]=1;
                mu[i*pri[j]]==0;
                break;
            }
        }
    }
    sum[0]=0;
    for (int i=1;i<=1000000ll;i++)
        sum[i]=sum[i-1]+phi[i];
    int T;
    scanf("%d",&T);
    while (T--)
    {
        // n=1000000;
        LL ans=0;
        scanf("%lld",&n);
        for (int i=n;i;i--)
        {
            a[i]=(n/i)*(n/i)*(n/i);
            ans+=a[i]*mu[i];
        }
        printf("%lld\n",ans+(sum[n]*2+1)*3+3);
    }

    return 0;
}

优化后

#include <cstdio>
#include <iostream>
#include <algorithm>
#include <ctime>
#include <cctype>
#include <cmath>
#include <string>
#include <cstring>
#include <stack>
#include <queue>
#include <list>
#include <vector>
#include <map>
#include <set>

#define sqr(x) ((x)*(x))
#define LL long long 
#define INF 0x3f3f3f3f
#define PI acos(-1.0)
#define eps 1e-10
using namespace std;
int mu[1000005];
int com[1000005];
int pri[1000005],pn=0;
int phi[1000005];
LL presum[1000005];
int musum[1000005];
int main()
{
	memset(com,0,sizeof com);
	presum[1]=0;
	mu[1]=1;
	phi[1]=0;
	for (int i=2;i<=1000000;i++)
 	{
  	if (com[i]==0)
  		{
   		pri[++pn]=i;
   		mu[i]=-1;
   		phi[i]=i-1;
  		}
  	for (int j=1;j<=pn&&pri[j]*i<=1000000;j++)
  	{
   	if (i%pri[j])
	   {
	    mu[i*pri[j]]=-mu[i];
	    com[i*pri[j]]=1;
	    phi[i*pri[j]]=phi[i]*(pri[j]-1);
	   }
	   else 
	   {
	    phi[i*pri[j]]=phi[i]*(pri[j]);
	    mu[i*pri[j]]=0;
	    com[i*pri[j]]=1;
	    break;
	   }
  	}
  	presum[i]=presum[i-1]+phi[i];
  	musum[i]=musum[i-1]+mu[i];
 	}
 int T;
 scanf("%d",&T);
 int a,b,c,d,k;
 while (T--)
 {
 	int n;
 	LL ans=0;
 	scanf("%d",&n);
 	int i;
 	for (i=1;i<=n;i++)
 		if ((n/i)==(n/(i+1))) break;
 		else 
 		ans+=(LL)(n/i)*(n/i)*(n/i)*mu[i];
 	for (int j=(n/i);j;j--)
 		ans+=(LL)(j)*(j)*(j)*(musum[n/(j)]-musum[n/(j+1)]);
 	ans+=(LL)presum[n]*6+6;
 	printf("%lld\n",ans);
 }
 return 0;
}