題面

LOJ

題解

戳這裏

#include<iostream>
#include<cstdio>
#include<cstdlib>
#include<cstring>
#include<cmath>
#include<algorithm>
using namespace std;
#define ll long long
#define MAX 222222
#define MOD 1000000007
ll n,Sqr,w[MAX];
ll pri[MAX],id1[MAX],id2[MAX],h[MAX],g[MAX],m;
bool
 
 zs[MAX];
int tot,sp[MAX];
void pre(int n)
{
    zs[1]=true;
    for(int i=2;i<=n;++i)
    {
        if(!zs[i])pri[++tot]=i,sp[tot]=(sp[tot-1]+i)%MOD;
        for(int j=1;j<=tot&&i*pri[j]<=n;++j)
        {
            zs[i*pri[j]]=true;
            if(i%pri[j]==0)break;
        }
    }
}
int
 
 S(ll x,int y)
{
    if(x<=1||pri[y]>x)return 0;
    int k=(x<=Sqr)?id1[x]:id2[n/x],ret=(g[k]-sp[y-1]-h[k]+y-1)%MOD;
    if(y==1)ret+=2;
    for(int i=y;i<=tot&&1ll*pri[i]*pri[i]<=x;++i)
    {
        ll t1=pri[i],t2=1ll*pri[i]*pri[i];
        for(int e=1;t2<=x;++e,t1=t2,t2*=pri[i])
            (ret+=((1ll
 
*S(x/t1,i+1)*(pri[i]^e)%MOD+(pri[i]^(e+1))%MOD)))%=MOD;
    }
    return ret;
}
int main()
{
    scanf("%lld",&n);Sqr=sqrt(n);
    pre(Sqr);
    for(ll i=1,j;i<=n;i=j+1)
    {
        j=n/(n/i);w[++m]=n/i;
        h[m]=(w[m]-1)%MOD;
        g[m]=(w[m]%MOD)*((w[m]+1)%MOD)%MOD;
        if(g[m]&1)g[m]=g[m]+MOD;g[m]/=2;g[m]--;
        if(w[m]<=Sqr)id1[w[m]]=m;
        else id2[j]=m;
    }
    for(int j=1;j<=tot;++j)
        for(int i=1;i<=m&&pri[j]*pri[j]<=w[i];++i)
        {
            int k=(w[i]/pri[j]<=Sqr)?id1[w[i]/pri[j]]:id2[n/(w[i]/pri[j])];
            (g[i]-=1ll*pri[j]*(g[k]-sp[j-1])%MOD)%=MOD;
            (h[i]-=h[k]-j+1)%=MOD;
        }
    int ans=S(n,1)+1;
    printf("%d\n",(ans+MOD)%MOD);
    return 0;
}

【LOJ6053】簡單的函數（min_25篩）