VLATTICE - Visible Lattice Points

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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

　

 

Description(题意)

有 N*N*N网格. 一个角落在 (0,0,0)，对顶角落是 (N,N,N). 问从(0,0,0)看有多少个格点是可见的？点 X从点Y可见，当且仅当，线段XY上没有其他的点。

Input:

第一行是测试数据个数T。接着有T行每行有一个整数 N.

Output :

输出T行，每行是对应的可见格点的个数。

Sample Input :

3

1

2

5

Sample Output :

7

19

175

 

Constraints :

T <= 50

1 <= N <= 1000000

 

Solution:

#include
     
      #include
      
       #ifdef WIN32#define LL "%I64d"#else#define LL "%lld"#endifusing namespace std;typedef long long ll;const int M=1e6+5;int n,m,T;ll sum[M];int tot,prime[M/3],mu[M];bool check[M];void sieve(){    n=1e6;mu[1]=1;    for(int i=2;i<=n;i++){        if(!check[i]) prime[++tot]=i,mu[i]=-1;        for(int j=1;j<=tot&&i*prime[j]<=n;j++){            check[i*prime[j]]=1;            if(!(i%prime[j])){mu[i*prime[j]]=0;break;}            else mu[i*prime[j]]=-mu[i];        }    }    for(int i=1;i<=n;i++) sum[i]=sum[i-1]+mu[i];}inline ll s2(int x){
   return 1LL*x*x;}inline ll s3(int x){
   return 1LL*x*x*x;}inline ll solve(int n){    ll ans=3;    for(int i=1,pos;i<=n;i=pos+1){        pos=n/(n/i);        ans+=s3(n/i)*(sum[pos]-sum[i-1]);        ans+=3*s2(n/i)*(sum[pos]-sum[i-1]);    }    return ans;}int main(){    sieve();    for(scanf("%d",&T);T--;){        scanf("%d",&n);        printf(LL"\n",solve(n));    }    return 0;}