SPOJ1007 VLATTICE - Visible Lattice Points

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<cstdio>
#include<iostream>
#ifdef WIN32
#define LL "%I64d"
#else
#define LL "%lld"
#endif
using 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;
}