F - 小清新数论

杜教筛能的，跑了1423ms，对上面公式中欧拉函数前n项和，欧拉函数前n项和进行杜教筛，然后套一个分块求解

#include<stdio.h>
#include<bits/stdc++.h>
#include<tr1/unordered_map>
#define INV2 499122177
using namespace std;
typedef long long ll;
const int N=1e7+20;
const int mod=998244353;
bool vis[N];
int mu[N],sum1[N];
long long phi[N],sum2[N];
int cnt,prim[N];
int e,e1;
tr1::unordered_map<long long,long long>w,w1;  //哈希 w用来求phi前缀和 w1用来求miu前缀和
void get(int maxn)
{
    phi[1]=mu[1]=1;
    for(int i=2;i<=maxn;i++)
    {
        if(!vis[i])
        {
            prim[++cnt]=i;
            mu[i]=-1;phi[i]=i-1;
        }
        for(int j=1;j<=cnt&&prim[j]*i<=maxn;j++)
        {
            vis[i*prim[j]]=1;
            if(i%prim[j]==0)
            {
                phi[i*prim[j]]=phi[i]*prim[j];
                break;
            }
            else mu[i*prim[j]]=-mu[i],phi[i*prim[j]]=phi[i]*(prim[j]-1);
        }
    }
    for(int i=1;i<=maxn;i++)sum1[i]=sum1[i-1]+mu[i],sum2[i]=(sum2[i-1]+phi[i])%mod;   //打一个maxn的phi前缀和表 和miu前缀和表
}
int djsmu(long long x)	// 求miu前缀和
{
    if(x<=10000000)return sum1[x];
    if(w[x])return w[x];
    int ans=1;
    for(long long l=2,r;l<=x;l=r+1)
    {
        r=x/(x/l);
        ans-=(r-l+1ll)*djsmu(x/l);
    }
    return w[x]=ans;
}
long long djsphi(long long x)	//求phi 前缀和
{
    if(x<=10000000)return sum2[x];
    if(w1[x])return w1[x];
    long long ans=x%mod*(x+1)%mod*INV2%mod;
    for(long long l=2,r;l<=x;l=r+1)
    {
        r=x/(x/l);
        ans=(ans-(r-l+1)%mod*djsphi(x/l)+mod)%mod;
    }
    while(ans<0)ans+=mod;
    return w1[x]=ans%mod;
}
int main(){
    get(10000000);
    ll n,r;
    scanf("%lld",&n);
    ll ans=0,res;
    for(ll l=1;l<=n;l=r+1){
        r=n/(n/l);
        res=(ll)(djsmu(r)-djsmu(l-1)+mod)%mod*((djsphi(n/l)%mod*(ll)2%mod-1+mod)%mod)%mod;
        ans=(ans+res+mod)%mod;
    }
    printf("%lld\n",ans);
    return 0;
}