POJ 3090 Visible Lattice Points(莫比乌斯反演)

#include<cstdio>
#include<cstring>
#include<algorithm>
#include<cmath>
using namespace std;
#define cl(a,b) memset(a,b,sizeof(a))
#define ll long long
#define pb push_back
#define gcd __gcd

const double EPS = 1e-8;
const int maxn = 1e3+1000;
const int inf  = 0x3f3f3f3f;
const double PI = acos(-1.0);

bool check[maxn];
int prime[maxn];
int mu[maxn];
void Moblus(){
cl(check,false);
mu[1]=1;int tot=0;
for(int i=2;i<maxn;i++){
if(!check[i]){
prime[tot++]=i;
mu[i]=-1;
}
for(int j=0;j<tot;j++){
if(i*prime[j]>maxn)break;
check[i*prime[j]]=true;
if(i%prime[j]==0){
mu[i*prime[j]]=0;break;
}
else mu[i*prime[j]]=-mu[i];
}
}
}
/*
int k;
//TLE 代码
ll cal(int n,int m){
n/=k;m/=k;
if(n>m)swap(n,m);
ll ans=0;
for(int i=1;i<=n;i++)ans+=(ll)mu[i]*(n/i)*(m/i);
return ans;
}*/
ll sum[maxn];
//分块优化 N/x值的种类个数不大于2*sqrt(N)
//计算[1,n],[1,m]互质的对数
ll cal(int n,int m){
if(n>m)swap(n,m);
ll ans=0;
for(int i=1,last=0;i<=n;i=last+1){
last=min(n/(n/i),m/(m/i));
ans+=(sum[last]-sum[i-1])*(n/i)*(m/i);
}
return ans;
}

int main(){

Moblus();
sum[0]=0;for(int i=1;i<maxn;i++)sum[i]=sum[i-1]+mu[i];
int T;scanf("%d",&T);
int cas=1;
while(T--){
int n;scanf("%d",&n);
printf("%d %d %lld\n",cas++,n,cal(n,n)+2);//gcd==1 and gcd ==0
}
return 0;
}