HYSBZ - 2818莫比乌斯反演

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题意很简洁不说了

题解：一开始我想直接暴力，复杂度是O（log（1e7）*sqrt(1e7)）算出来是2e9，可能会复杂度爆炸，但是我看时限是10s，直接大力莽了一发暴力，没想到就过了= =

就是先打出1e7的素数表，然后挨个算即可

//#pragma comment(linker, "/stack:200000000")

//#pragma GCC optimize("Ofast,no-stack-protector")

//#pragma GCC target("sse,sse2,sse3,ssse3,sse4,popcnt,abm,mmx,avx,tune=native")

//#pragma GCC optimize("unroll-loops")

#include<bits/stdc++.h>

#define fi first

#define se second

#define mp make_pair

#define pb push_back

#define pi acos(-1.0)

#define ll long long

#define mod 1000000007

#define C 0.5772156649

#define ls l,m,rt<<1

#define rs m+1,r,rt<<1|1

#define pil pair<int,ll>

#define pii pair<int,int>

#define ull unsigned long long

#define base 1000000000000000000

#define fio ios::sync_with_stdio(false);cin.tie(0)

using namespace std;

const double g=10.0,eps=1e-;

const int N=+,maxn=+,inf=0x3f3f3f3f;

int mu[N],prime[N],sum[N];

bool mark[N];

int cnt;

void init()

{

    mu[]=;

    cnt=;

    for(int i=;i<N;i++)

    {

        if(!mark[i])prime[++cnt]=i,mu[i]=-;

        for(int j=;j<=cnt;j++)

        {

            int t=i*prime[j];

            if(t>N)break;

            mark[t]=;

            if(i%prime[j]==){mu[t]=;break;}

            else mu[t]=-mu[i];

        }

    }

    for(int i=;i<N;i++)sum[i]=sum[i-]+mu[i];

}

int main()

{

    init();

    int n;

    scanf("%d",&n);

    ll ans=;

    for(int i=;i<=cnt;i++)

    {

        int te=n/prime[i];

        for(int j=,last;j<=te;j=last+)

        {

            last=te/(te/j);

            ans+=(ll)(sum[last]-sum[j-])*(te/j)*(te/j);

        }

    }

    printf("%lld\n",ans);

    return ;

}

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