正题

P2568

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题目大意

求满足$1\le x,y\le n$且$gcd(x,y)=prime$的数对$(x,y)$的个数

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解题思路

题目即求

$$\sum _{i=1}^n\sum _{j=1}^n[gcd(i,j)=prime]$$

可以考虑先枚举该prime，那么有

$$\begin{gathered} \sum _{p\in prime}^n\sum _{i=1}^{n/p}\sum _{j=1}^{n/p}[gcd(i,j)=1] \\ \sum _{p\in prime}^n\sum _{i=1}^{n/p}\sum _{j=1}^{n/p}\sum _{d|i,d|j}\mu (d) \\ \sum _{p\in prime}^n\sum _{d=1}^{n/p}\mu (d)\times \lfloor \frac{n}{d}\rfloor \times \lfloor \frac{n}{d}\rfloor \end{gathered}$$

时间复杂度不太会证，看dalao证的是 $O(n/logn)$的

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code

#include<cstdio>
#include<cstring>
#include<iostream>
#include<algorithm>
#define ll long long
#define N 10000010
using namespace std;
ll n,ans,w,p[N],mu[N],prime[N];
void work()
{p[1]=mu[1]=1;for(ll i=2;i<=10000000;++i){if(!p[i]){prime[++w]=i;mu[i]=-1;}for(ll j=1;j<=w&&i*prime[j]<=10000000;++j){p[i*prime[j]]=1;if(i%prime[j]==0){mu[i*prime[j]]=0;break;}else mu[i*prime[j]]=-mu[i];}}for(ll i=2;i<=10000000;++i)mu[i]+=mu[i-1];return;
}
ll get(ll n)
{ll sum=0;for(ll l=1,r=0;l<=n;l=r+1){r=n/(n/l);sum+=(mu[r]-mu[l-1])*(n/l)*(n/l);}return sum;
}
int main()
{scanf("%lld",&n);work();for(ll i=1;i<=w;++i)if(prime[i]<=n)ans+=get(n/prime[i]);else break;printf("%lld",ans);return 0;
}