【LOJ6053】简单的函数（min_25筛）

题面

LOJ

题解

戳这里

#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;

}