题目链接
一定要记得随时取模和%mmh学长。

#include<iostream>
#include<cstdio>
#include<algorithm>
#include<cstring>
#include<cmath>
#include<string>
#include<queue>
#include<map>
#include<unordered_map>
#define ll long long
#define llu unsigned ll
using namespace std;
const int maxn=200100;
const int mod=1e9+7;
ll p[maxn],id1[maxn],id2[maxn];
ll g[maxn],h[maxn],sum[maxn],sum2[maxn],wi[maxn];
int cnt;
bool ha[maxn];
ll n,inv2,inv6,sq;

ll mypow(ll a,ll b)
{
    ll ans=1;
    while(b)
    {
        if(b&1) ans=ans*a%mod;
        a=a*a%mod;
        b>>=1;
    }
    return ans;
}

void Prime(ll n)
{
   ha[1]=true;
   for(int i=2;i<=n;i++)
   {
       if(!ha[i])
       {
           p[++cnt]=i;
           sum[cnt]=(sum[cnt-1]+i)%mod;
           sum2[cnt]=(sum2[cnt-1]+1ll*i*i)%mod;
       }
       for(int j=1;j<=cnt&&p[j]*i<=n;j++)
       {
           ha[i*p[j]]=true;
           if(i%p[j]==0) break;
       }
   }
}

int getid(ll m)
{
    if(m<=sq) return id1[m];
    else return id2[n/m];
}

ll s(ll n,ll j)
{
    if(n<=1||p[j]>n) return 0;
    int k=getid(n);
    ll res=(((g[k]-sum2[j-1])-(h[k]-sum[j-1]))%mod+mod)%mod;

    for(int i=j;i<=cnt&&p[i]*p[i]<=n;i++)
    {
        ll p1=p[i],p2=p[i]*p[i];
        for(int e=1;p2<=n;e++,p1=p2,p2=p2*p[i])
        {
            res=(res+s(n/p1,i+1)*(p1%mod)%mod*((p1-1)%mod)%mod+p2%mod*((p2-1)%mod)%mod)%mod;
        }
    }
    return res;
}

ll getans1(ll n)
{
    n%=mod;
    return (n*(n+1)%mod*(2*n+1)%mod*inv6%mod-1+mod)%mod;
}

ll getans2(ll n)
{
    n%=mod;
    return (n*(n+1)%mod*inv2%mod-1+mod)%mod;
}

int main(void)
{
    inv2=mypow(2,mod-2);
    inv6=mypow(6,mod-2);
    scanf("%lld",&n);
    sq=sqrt(n*1.0);
    Prime(sq);

    int tot=0;
    for(ll l=1,r;l<=n;l=r+1)
    {
        r=n/(n/l);
        wi[++tot]=n/l;
        if(wi[tot]<=sq) id1[wi[tot]]=tot;
        else id2[n/wi[tot]]=tot;
        g[tot]=getans1(wi[tot]);
        h[tot]=getans2(wi[tot]);
    }
    for(int j=1;j<=cnt;j++)
    {
        for(int i=1;i<=tot&&p[j]*p[j]<=wi[i];i++)
        {
            int k=getid(wi[i]/p[j]);
            g[i]=((g[i]-p[j]*p[j]%mod*(g[k]-sum2[j-1])%mod)%mod+mod)%mod;
            h[i]=((h[i]-p[j]*(h[k]-sum[j-1])%mod)%mod+mod)%mod;
        }
    }
    ll ans=s(n,1);
    printf("%lld\n",(ans+1)%mod);
    return 0;
}