dtoj3571. 乌(mi)

我有一段代码：

输入N，求G(N)的值.

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sol

$sumlimits_{i=1}^{n} sumlimits_{j=1}^{n/i} h[i imes j]+=[gcd(i,j)==1] $

$sumlimits_{d} sumlimits_{i=1}^{n/d} sumlimits_{j=1}^{n/id} h[ijd^2]+=mu(d)$
由于我们求$sum h$
$calc(x)=sumlimits_{i=1}^{n} sumlimits_{j=1}^{n/i} 1$
$sumlimits_{d} sumlimits_{i=1}^{n/d} sumlimits_{j=1}^{n/id} mu(d)*calc(n/d^2)$

前面d的枚举是$sqrt(n)$的 ,calc整数分块
那么效率是$sqrt{n} +sqrt{n/4}+sqrt{n/9}...$
感觉是$ sqrt{n} ln(n)$

#include<cstdio>
#include<iostream>
#include<cstdlib>
#include<cstring>
#include<algorithm>
#include<cmath>
#define maxn 100005
#define mod 998244353
using namespace std;
int n,N,pri[maxn],flag[maxn],tot,mu[maxn],ans;
int calc(int x){
    int sum=0;
    for(int i=1,nex;i<=x;i=nex+1){
        nex=x/(x/i);
        sum=(sum+1LL*(nex-i+1)*(x/i)%mod)%mod;
    }
    return sum;
}
int main(){
    cin>>n;N=sqrt(n)+1;mu[1]=1;
    for(int i=2;i<=N;i++){
        if(!flag[i]){pri[++tot]=i;mu[i]=-1;}
        for(int j=1;j<=tot&&i*pri[j]<=N;j++){
            flag[i*pri[j]]=1;
            if(i%pri[j]==0){mu[i*pri[j]]=0;break;}
            mu[i*pri[j]]=-mu[i];
        }
    }
    for(int i=1;i<=N;i++){
        if(mu[i]!=0){
            ans=(ans+mu[i]*calc(n/(i*i)))%mod;
        }
    }
    ans=(ans+mod)%mod;
    cout<<ans<<endl;
    return 0;
}