P2568 GCD（莫比乌斯反演）
作者：互联网
题目传送门
本题题意转化成为：
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\displaystyle\sum_{i=1}^{n}\displaystyle\sum_{j=1}^{n}[ gcd(i,j)==p]
i=1∑nj=1∑n[gcd(i,j)==p]
这道题用莫比乌斯式子推其实也挺简单的说…………欧拉函数的那个还理解了半天我真菜
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step:1提取公因式
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\displaystyle\sum_{p=1}^{p∈prime,p<=n}\displaystyle\sum_{i=1}^{[n/p]}\displaystyle\sum_{j=1}^{[n/p]}ε (gcd(i,j))
p=1∑p∈prime,p<=ni=1∑[n/p]j=1∑[n/p]ε(gcd(i,j))
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step:2利用迪利克雷卷积进行反演
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\displaystyle\sum_{p=1}^{p∈prime,p<=n}\displaystyle\sum_{i=1}^{[n/p]}\displaystyle\sum_{j=1}^{[n/p]}\displaystyle\sum_{k|i,k|j}^{}μ(k)
p=1∑p∈prime,p<=ni=1∑[n/p]j=1∑[n/p]k∣i,k∣j∑μ(k)
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step:3交换枚举顺序
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\displaystyle\sum_{p=1}^{p∈prime,p<=n}\displaystyle\sum_{k=1}^{[n/p]}\displaystyle\sum_{i=1}^{[n/(p*k)]}\displaystyle\sum_{j=1}^{[n/(p*k)]}μ(k)
p=1∑p∈prime,p<=nk=1∑[n/p]i=1∑[n/(p∗k)]j=1∑[n/(p∗k)]μ(k)
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\displaystyle\sum_{p=1}^{p∈prime,p<=n}\displaystyle\sum_{k=1}^{[n/p]}μ(k)*(n/(p*k))^2
p=1∑p∈prime,p<=nk=1∑[n/p]μ(k)∗(n/(p∗k))2
然后利用我们的整除分块就可以愉快的实现我们的代码了
#include<iostream>
#include<vector>
using namespace std;
typedef long long ll;
const ll MAXN=1e7+10;
//莫比乌斯函数
ll mu[MAXN];
bool isnp[MAXN];
vector<ll> primes;
void init(ll n)
{
mu[1] = 1;
for (ll i = 2; i <= n; i++)
{
if (!isnp[i])
primes.push_back(i), mu[i] = -1; // 质数为-1
for (ll p : primes)
{
if (p * i > n)
break;
isnp[p * i] = 1;
if (i % p == 0)
{
mu[p * i] = 0; // 有平方因数为0
break;
}
else
mu[p * i] = mu[p] * mu[i]; // 互质，利用积性函数性质
}
}
for(ll i=1;i<=n;i++){
mu[i]+=mu[i-1];
}
}
int main(){
ll n,ans=0;
init(10000000);
cin>>n;
for(ll i=0;primes[i]<=n;i++){
//cout<<primes[i]<<endl;
ll u=n/primes[i];
ll l=1,r;
for(ll l=1,r;l<=u;l=r+1){
r=u/(u/l);
ans+=(mu[r]-mu[l-1])*(u/l)*(u/l);
}
}
cout<<ans<<endl;
}
标签：mu,GCD,ll,P2568,反演,step,MAXN,isnp,primes 来源： https://blog.csdn.net/m0_51841071/article/details/116399741