[SPOJ VLATTICE]Visible Lattice Points 数论 莫比乌斯反演
7001. Visible Lattice PointsProblem 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 :
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，求和即可
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 可以省去一次次计算，直接按mu的前缀和来处理
则我们就统计相同(n/i)的值是否出现两次，如果出现两次那么我们就开始按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;
}