[BZOJ2818]Gcd（莫比乌斯反演）

题目描述

传送门

题解

∑i=1n∑j=1n∑d=1prime[0][gcd(i,j)=prime[d]]

=∑i=1n∑j=1n∑d=1prime[0][gcd(iprime[d],jprime[d])=1]

=∑d=1prime[0]∑i=1nprime[d]∑j=1nprime[d][gcd(i,j)=1]

=∑d=1prime[0]∑i=1nprime[d]∑j=1nprime[d]∑t|gcd(i,j)μ(t)

=∑d=1prime[0]∑i=1nprime[d]∑j=1nprime[d]∑t=1n[t|i][t|j]μ(t)

=∑d=1prime[0]∑t=1n∑i=1nprime[d][t|i]∑j=1nprime[d][t|j]μ(t)

=∑d=1prime[0]∑t=1nnprime[d]tnprime[d]tμ(t)

枚举质数是必须要枚举的，后面的那一坨根n分块，预处理mu的前缀和。

代码

#include<algorithm>
#include<iostream>
#include<cstring>
#include<cstdio>
using namespace std;
#define LL long long

const int max_n=1e7+5;

LL n;
LL ans;
LL p[max_n],prime[max_n],mu[max_n];

inline void get_mu(int n){
mu[1]=1;
for (int i=2;i<=n;++i){
if (!p[i]){
prime[++prime[0]]=i;
mu[i]=-1;
}
for (int j=1;j<=prime[0]&&i*prime[j]<=n;++j){
p[i*prime[j]]=1;
if (i%prime[j]==0){
mu[i*prime[j]]=0;
break;
}
else mu[i*prime[j]]=-mu[i];
}
mu[i]+=mu[i-1];
}
}

int main(){
scanf("%lld",&n);
get_mu(n);
LL j;
for (LL d=1;d<=prime[0]&&prime[d]<=n;++d){
LL N=(LL)n/prime[d];
for (LL i=1;i<=N;i=j+1){
j=min(N,N/(N/i));
ans+=(N/i)*(N/i)*(mu[j]-mu[i-1]);
}
}
printf("%lld\n",ans);
}