loj6053(min25筛)

题目链接

https://loj.ac/problem/6053

题解

min25筛模板题,做这一类题的时候关键要把 $f(p)$ 用完全积性函数表示

在这题中,$f(p)=p\oplus1$ ,而只有 $p=2$ 时, $f(p)=3$ ,对其他情况 $f(p)=p-1$ ,那么可以先设 $f(p)=p-1$ ,再把 $2$ 的影响去掉就可以了。。




代码

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/**
*         ┏┓    ┏┓
*         ┏┛┗━━━━━━━┛┗━━━┓
*         ┃       ┃  
*         ┃   ━    ┃
*         ┃ >   < ┃
*         ┃       ┃
*         ┃... ⌒ ...  ┃
*         ┃ ┃
*         ┗━┓ ┏━┛
*          ┃ ┃ Code is far away from bug with the animal protecting          
*          ┃ ┃ 神兽保佑,代码无bug
*          ┃ ┃           
*          ┃ ┃       
*          ┃ ┃
*          ┃ ┃           
*          ┃ ┗━━━┓
*          ┃ ┣┓
*          ┃ ┏┛
*          ┗┓┓┏━━━━━━━━┳┓┏┛
*           ┃┫┫ ┃┫┫
*           ┗┻┛ ┗┻┛
*/

#include<cstdio>
#include<cstring>
#include<algorithm>
#include<iostream>
#include<queue>
#include<map>
#include<stack>
#include<set>
#include<bitset>
#include<cmath>
#include<assert.h>
#define inc(i,l,r) for(int i=l;i<=r;i++)
#define dec(i,l,r) for(int i=l;i>=r;i--)
#define link(x) for(edge *j=h[x];j;j=j->next)
#define mem(a) memset(a,0,sizeof(a))
#define ll long long
#define eps 1e-8
#define succ(x) (1<<x)
#define lowbit(x) (x&(-x))
#define sqr(x) ((x)*(x))
#define mid (x+y>>1)
#define NM 2000005
#define nm 128
#define pi 3.1415926535897931
using namespace std;
const ll inf=1e9+7;
ll read(){
ll x=0,f=1;char ch=getchar();
while(!isdigit(ch)){if(ch=='-')f=-1;ch=getchar();}
while(isdigit(ch))x=x*10+ch-'0',ch=getchar();
return f*x;
}





int tot,m;
ll w[NM],pre[NM],f[NM],h[NM],prime[NM],n;
bool v[NM];

inline void reduce(ll&x){x+=x>>63&inf;}
inline int id(ll x){return x<=m/2?m-x+1:n/x;}

void init(){
m=sqrt(n);
inc(i,2,m)if(!v[i]){
prime[++tot]=i;
for(int j=i<<1;j<=m;j+=i)v[j]++;
}
inc(i,1,tot)pre[i]=(pre[i-1]+prime[i])%inf;
inc(i,1,m)w[i]=n/i;
while(w[m]>1)w[m+1]=w[m]-1,m++;
}

ll S(ll n,int m){
if(n<prime[m])return 0;
ll ans=(f[id(n)]-pre[m-1]+inf)%inf;
for(int i=m;i<=tot&&prime[i]*prime[i]<=n;i++){
ll t=prime[i];
for(int j=1;t*prime[i]<=n;t*=prime[i],j++)
reduce(ans+=(prime[i]^j)*S(n/t,i+1)%inf-inf),
reduce(ans+=(prime[i]^(j+1))-inf);
}
return ans;
}

int main(){
n=read();
init();
inc(i,1,m)f[i]=w[i]%inf*((w[i]+1)%inf)%inf*(inf+1)/2%inf-1,h[i]=(w[i]-1)%inf;
inc(j,1,tot){
inc(i,1,m)if(w[i]>=prime[j]*prime[j]){
int k=id(w[i]/prime[j]);
reduce(f[i]-=prime[j]*(f[k]-pre[j-1]+inf)%inf);
reduce(h[i]-=h[k]-j+1);
}else break;
}
inc(i,1,m-1)f[i]+=2;
inc(i,1,m)reduce(f[i]-=h[i]);
inc(i,1,tot)pre[i]=(pre[i-1]+(prime[i]^1))%inf;
printf("%lld\n",S(n,1)+1);
return 0;
}