I have a piece of code:
Input N, the evaluation G (N) is.
sol
$\sum\limits_{i=1}^{n} \sum\limits_{j=1}^{n/i} h[i \times j]+=[gcd(i,j)==1] $
$\sum\limits_{d} \sum\limits_{i=1}^{n/d} \sum\limits_{j=1}^{n/id} h[ijd^2]+=\mu(d)$
由于我们求$\sum h$
$calc(x)=\sum\limits_{i=1}^{n} \sum\limits_{j=1}^{n/i} 1$
$\sum\limits_{d} \sum\limits_{i=1}^{n/d} \sum\limits_{j=1}^{n/id} \mu(d)*calc(n/d^2)$
The foregoing enumerations are d $ sqrt (n) $ a, calc integer block
then the efficiency is $ \ sqrt {n} + \ sqrt {n / 4} + \ sqrt {n / 9} ... $
feeling $ \ 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; }