洛谷P2568 GCD
\(Solution:\)
根据题意我们可以推出以下式子
\(\sum\limits_{p \in prime}\;\sum\limits_{i=1}^n\;\sum\limits_{j=1}^n \;[\gcd(i,j)=p]\)
\(\because \gcd(a*x,a*y)=p=\gcd(x,y)=1\)
\(\therefore\)可得出下式
\(\sum\limits_{p \in prime}\;\sum\limits_{i=1}^{\lfloor\frac{n}{p}\rfloor}\;\sum\limits_{j=1}^{\lfloor\frac{n}{p}\rfloor}\;[\gcd(i,j)=1]\)
又\(\because\)是有序数对,但是存在两数都是素数的情况
\(\therefore\)可得出下式
\(\sum\limits_{p \in prime}\;\left(\sum\limits_{i=1}^{\lfloor\frac{n}{p}\rfloor}\;\left(2\sum\limits_{j=1}^i[\gcd(i,j)=1]\right)-1\right)\)
\(\because \gcd(i,j)=1\)可以用欧拉函数求解
\(\therefore\)可得出下式
\(\sum\limits_{p \in prime}\left(2\sum\limits_{i=1}^{\lfloor\frac{n}{p}\rfloor}\phi(i)-1\right)\)
最后求完这些东西,再加上一个前缀和优化即可
\(Code:\)
#include<bits/stdc++.h>
#define int long long
using namespace std;
inline int read()
{
int x=0,f=1;char c=getchar();
while(c<'0' || c>'9'){if(c=='-') f=0;c=getchar();}
while(c>='0' && c<='9') x=(x<<3)+(x<<1)+(c^48),c=getchar();
return f?x:-x;
}
const int N=1e7+10,M=1e6+10;
int n,cnt,p[M],phi[N],sum[N],ans;
bool vis[N];
void Euler_seive()
{
phi[1]=1;
for(int i=2;i<=n;i++)
{
if(!vis[i]){p[++cnt]=i;phi[i]=i-1;}
for(int j=1;j<=cnt && i*p[j]<=n;j++)
{
vis[i*p[j]]=1;
if(i%p[j]==0)
{
phi[i*p[j]]=phi[i]*p[j];
break;
}
else phi[i*p[j]]=phi[i]*phi[p[j]];
}
}
}
signed main()
{
n=read();
Euler_seive();
for(int i=1;i<=n;i++) sum[i]=sum[i-1]+phi[i];
for(int i=1;i<=cnt;i++) ans+=2*sum[n/p[i]]-1;
printf("%lld",ans);
return 0;
}