三维晶格可见点数计算-优快云博客

本文链接：https://blog.youkuaiyun.com/evilwind2000/article/details/98750303

本文介绍了一种使用莫比乌斯反演算法解决三维晶格中从一角看另一角可见点数的问题，通过整除分块和前缀和技巧，实现了高效计算。文章包含完整的C++代码实现。

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Consider a NNN 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
空间和平面上的点都可用莫比乌斯反演解决
懒得写公式了。。。参考博客：https://www.cnblogs.com/shenben/p/6748677.html

#include<bits/stdc++.h>
#define N 1000005
#define ll long long
using namespace std;
int p[N],check[N],tot;
int mu[N];
ll sum[N];
int T,n,m,d;
ll ans;
void init()              //算出莫比乌斯函数和前缀和 
{
	mu[1]=1;
	n=1e6;
	for(int i=2;i<=n;i++)
	{
		if (!check[i])
		{
			p[++tot]=i;
			mu[i]=-1;
		}
		for (int j=1;j<=tot && p[j]*i<=n;j++)
		{
			check[i*p[j]]=1;
			if (i%p[j]==0)
			{
				mu[i*p[j]]=0;
				break;	
			}
			else mu[i*p[j]]=-mu[i];
		}
	}
	for(int i=1;i<=n;i++) 
	sum[i]=(ll)mu[i]+sum[i-1];	
}
ll calc(int n)               //整除分块 
{
	ll ret=3;
	for (int L=1,R;L<=n;L=R+1)
	{
		R=n/(n/L);	
		ret+=(1ll)*(sum[R]-sum[L-1])*((1ll)*(n/L)*(n/L)*(n/L)+3ll*(n/L)*(n/L));
	}
	return ret;
} 
int main()
{
	init();
	scanf("%d",&T);
	while (T--)
	{
		scanf("%d",&n);
		ans=calc(n);
		printf("%lld\n",ans);
	}
	return 0;
}