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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-12;
const int N=10000000+10,maxn=400000+10,inf=0x3f3f3f3f;

int mu[N],prime[N],sum[N];
bool mark[N];
int cnt;
void init()
{
    mu[1]=1;
    cnt=0;
    for(int i=2;i<N;i++)
    {
        if(!mark[i])prime[++cnt]=i,mu[i]=-1;
        for(int j=1;j<=cnt;j++)
        {
            int t=i*prime[j];
            if(t>N)break;
            mark[t]=1;
            if(i%prime[j]==0){mu[t]=0;break;}
            else mu[t]=-mu[i];
        }
    }
    for(int i=1;i<N;i++)sum[i]=sum[i-1]+mu[i];
}
int main()
{
    init();
    int n;
    scanf("%d",&n);
    ll ans=0;
    for(int i=1;i<=cnt;i++)
    {
        int te=n/prime[i];
        for(int j=1,last;j<=te;j=last+1)
        {
            last=te/(te/j);
            ans+=(ll)(sum[last]-sum[j-1])*(te/j)*(te/j);
        }
    }
    printf("%lld
",ans);
    return 0;
}
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