题面
题解
这种题目就是推倒推倒
\[\sum_{i=1}^n \gcd(i,n)=\sum_{i|n}i\sum_{j=1}^n[\gcd(j,n)=i]\]
\[\sum_{i=1}^n \gcd(i,n)=\sum_{i|n}i\sum_{j=1}^{\frac{i}{n}}[\gcd(j,\frac{i}{n})=1]\]
\[\sum_{i=1}^n \gcd(i,n)=\sum_{i|n}i\varphi({\frac{i}{n}})\]
然后暴力就行了
//minamoto
#include<bits/stdc++.h>
#define R register
#define ll long long
#define fp(i,a,b) for(R int i=a,I=b+1;i<I;++i)
#define fd(i,a,b) for(R int i=a,I=b-1;i>I;--i)
#define go(u) for(int i=head[u],v=e[i].v;i;i=e[i].nx,v=e[i].v)
using namespace std;
const int N=1e5+5;
bitset<N>vis;int p[N],phi[N],c[N],v[N],top,n,m,t,sqr;ll res;
void init(int n){
phi[1]=1;
fp(i,2,n){
if(!vis[i])p[++m]=i,phi[i]=i-1;
for(R int j=1;j<=m&&1ll*i*p[j]<=n;++j){
vis[i*p[j]]=1;
if(i%p[j]==0){phi[i*p[j]]=phi[i]*p[j];break;}
phi[i*p[j]]=phi[i]*(p[j]-1);
}
}
}
int Phi(int n){
if(n<=sqr)return phi[n];
int res=n;
for(R int i=1;i<=m&&p[i]<=n;++i)if(n%p[i]==0){
while(n%p[i]==0)n/=p[i];
res=res/p[i]*(p[i]-1);
}
if(n!=1)res=res/n*(n-1);
return res;
}
void dfs(int pos,int now){
if(pos==top+1)return res+=1ll*now*Phi(n/now),void();
for(int i=0;i<=c[pos];++i,now*=v[pos])dfs(pos+1,now);
}
int main(){
// freopen("testdata.in","r",stdin);
scanf("%d",&n),init(sqr=N-5);
t=n;
fp(i,1,m)if(t%p[i]==0){
v[++top]=p[i];
while(t%p[i]==0)++c[top],t/=p[i];
}
if(t!=1)v[++top]=t,c[top]=1;
dfs(1,1);
printf("%lld\n",res);
return 0;
}