Mobius an anti-drill exercises

The meaning of problems

$n<=1e9$ , the answer to $2^{32}$ Take the film
portion minutes $n<=1e6$, $n<=1e7$
first discovered $mu(i*j)=mu(i)*mu(j)*[gcd(i,j)==1]$
and then lost the final inversion $[gcd(i,j)==1]$ can be $O (nlogn)$
for $n<=1e6$ , it can be done directly
, but this approach constant is small, so add some constant optimization, it can $n<=1e7$

#include<bits/stdc++.h>
using namespace std;
#define int unsigned int//无符号整型速度很快
const int maxn=1e7+5;
inline int read(){
	char c=getchar();int t=0,f=1;
	while(!isdigit(c)){if(c=='-')f=-1;c=getchar();}
	while(isdigit(c)){t=(t<<3)+(t<<1)+(c^48);c=getchar();}
	return t*f;
}
int mu[maxn],p[maxn],vis[maxn],n,cnt;
void get(){
	mu[1]=1;
	for(int i=2;i<=n;i++){
		if(!vis[i]){
			p[++cnt]=i;mu[i]=-1;
		}
		for(int j=1;i*p[j]<=n&&j<=cnt;j++){
			vis[i*p[j]]=1;
			mu[i*p[j]]=mu[i]*mu[p[j]];
			if(i%p[j]==0){
				mu[i*p[j]]=0;
				break;
			}
		}
	}
}
int alfa[maxn];
signed main(){
	//freopen("T2.in","r",stdin);
	//freopen("T2.out","w",stdout);
	n=read();
	get();
	for(int i=n;i>=1;i--){
		int tmp=n/i;
		if(i*2<=n)//这里是常数优化
		alfa[i]=alfa[i*2];
		for(int j=1;j<=tmp;j+=2)//注意这里的步长
		alfa[i]=alfa[i]+mu[i*j];
	}
	int ans=0;
	for(int i=1;i<=n;i++){
		int tmp=mu[i]*alfa[i]*alfa[i];
		ans=ans+tmp;
	}
	printf("%u\n",ans);
	return 0;
}