传送门
题解：设p为素数 ，则gcd(x/p,y/p)=1也就是说求 x／p以及 y／p的欧拉函数。欧拉筛+前缀和就可以解决
#include <iostream> #include <cstdio> #include <cmath> #include <algorithm> #include <map> #include <queue> #include <vector> #include <cstring> #include <iomanip> #include <set> #include<ctime> //#include<unordered_map> //CLOCKS_PER_SEC #define se second #define fi first #define ll long long #define lson l,m,rt<<1 #define rson m+1,r,rt<<1|1 #define Pii pair<int,int> #define Pli pair<ll,int> #define ull unsigned long long #define pb push_back #define fio ios::sync_with_stdio(false);cin.tie(0) const int N=1e7+10; const ull base=163; const int INF=0x3f3f3f3f; using namespace std; bool ispri[N]; ll pri[N],phi[N]; int tot=0; void getphi(int x) {memset(ispri,1,sizeof(ispri));ispri[1]=false;phi[1]=1;for(int i=2;i<=x;i++){if(ispri[i]){pri[++tot]=i;phi[i]=i-1;}for(int j=1;j<=tot&&i*pri[j]<=x;j++){ispri[i*pri[j]]=false;if(i%pri[j]==0){phi[i*pri[j]]=phi[i]*pri[j];break;}phi[i*pri[j]]=phi[i]*(pri[j]-1);}} } ll sum[N]; int main(){int n;cin>>n;getphi(n);ll ans=0;for(int i=1;i<=n;i++)sum[i]=phi[i]+sum[i-1];for(int i=1;i<=tot;++i){ans+=sum[n/pri[i]]*2-1;}cout<<ans<<endl;return 0; }
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