參考陳牧歌在apio2018的講課課件

#include <iostream>
#include <cstring>
#include <cstdio>
using namespace std;
typedef long long ll;
ll n;
int pri[4641590], cnt, phi[4641590], msh[1544665][2];
bool isp[4641590];
const int mod=1000000007;
int getHash(ll x){
    int p=x%1544657;
    while(msh[p][0] && msh[p][0
 
]!=x)    p = (p + 1) % 1544657;
    return p;
}
int getPhi(ll x){
    if(x<=4641581)  return phi[x];
    int p=getHash(x);
    if(msh[p][0])   return msh[p][1];
    msh[p][0] = x;
    ll lst, re=0;
    for(ll i=2; i<=x; i=lst+1){
        lst = x / (x / i);
        ll tmp=(ll)getPhi(x/i) * (lst-i+1) % mod;
        re = (re + tmp) % mod;
    }
    x %= mod;
    re = ((ll)x * (x + 1
 
) % mod * 500000004 % mod - re + mod) % mod;
    msh[p][1] = re;
    return re;
}
int main(){
    memset(isp, true, sizeof(isp));
    isp[0] = isp[1] = false;
    phi[1] = 1;
    for(int i=2; i<=4641581; i++){
        if(isp[i])  pri[++cnt] = i, phi[i] = i - 1;
        for(int j=1; j<=cnt && i*pri[j]<=4641581
 
; j++){
            isp[i*pri[j]] = false;
            if(i%pri[j]==0){
                phi[i*pri[j]] = phi[i] * pri[j];
                break;
            }
            else    phi[i*pri[j]] = phi[i] * (pri[j] - 1);
        }
    }
    for(int i=2; i<=4641581; i++)
        phi[i] = (phi[i-1] + phi[i]) % mod;
    cin>>n;
    cout<<getPhi(n)<<endl;
    return 0;
}

51nod1239 歐拉函數之和