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+;
int n,m,T;ll sum[M];
int tot,prime[M/],mu[M];bool check[M];
void sieve(){
    n=1e6;mu[]=;
    for(int i=;i<=n;i++){
        if(!check[i]) prime[++tot]=i,mu[i]=-;
        for(int j=;j<=tot&&i*prime[j]<=n;j++){
            check[i*prime[j]]=;
            if(!(i%prime[j])){mu[i*prime[j]]=;break;}
            else mu[i*prime[j]]=-mu[i];
        }
    }
    for(int i=;i<=n;i++) sum[i]=sum[i-]+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=;
    for(int i=,pos;i<=n;i=pos+){
        pos=n/(n/i);
        ans+=s3(n/i)*(sum[pos]-sum[i-]);
        ans+=*s2(n/i)*(sum[pos]-sum[i-]);
    }
    return ans;
}
int main(){
    sieve();
    for(scanf("%d",&T);T--;){
        scanf("%d",&n);
        printf(LL"\n",solve(n));
    }
    return ;
}