[Spoj vlattice] visible lattice points Number Theory Mobius Inversion

7001. Visible lattice pointsproblem code: vlattice

Consider a n * n lattice. one corner is at (0, 0) and the opposite one is at (N, N, N ). how many lattice points are visible from corner at (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

Theme

Given a cube of N * n, each integer point has a lamp (abstract understanding) except (0, 0, 0 ), how many lights can be seen? (the covered lights are not counted)

Solutions

Typical application of Mobius Inversion/rejection Principle

The principle of rejection is that all three vertices can be multiplied by the number of vertices divisible by K by the Mobius coefficient, and the sum is enough.

The number of all three vertices that can be divisible by K is floor (N/I) ^ 3.

Note that the maximum data volume is 1000000 direct linear processing possible TLE

However, some results of (N/I) in I> (n/2) below can all be calculated once and again, and are processed by the Mu prefix and

Then, we will count whether the same (N/I) value appears twice. If it appears twice, we will start to process it by prefix and method.

6200 MS not optimized

After optimization, 490 Ms

Code

Before Optimization

#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 100000007llusing 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;}

After Optimization

#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-10using 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;}

[Spoj vlattice] visible lattice points Number Theory Mobius Inversion