LOJ #6053. 简单的函数

$Min$_$25$筛模版题

为什么泥萌常数都那么小啊$ QAQ$

传送门：Here

---

题意：

$ f(1)=1$
$ f(p^c)=p⊕c（p 为质数，⊕ 表示异或）$
$ f(ab)=f(a)f(b)（a 与 b 互质）$

求$ \sum\limits_{i=1}^n f(i)$

---

$ solution:$

显然有$ f(P_i)=P_i-1+2*[P_i=2]$

暂时忽略$ P_i=2$的情况求出质数贡献

然后再把答案$ +2$即可

---

$ my \ code: $

#include<ctime>
#include<cmath>
#include<cstdio>
#include<cstring>
#include<iostream>
#include<algorithm>
#include<queue>
#define p 1000000007
#define M 200010
#define rt register int
#define ll long long
using namespace std;
inline ll read(){
    ll x = ; char zf = ; char ch = getchar();
    while (ch != '-' && !isdigit(ch)) ch = getchar();
    if (ch == '-') zf = -, ch = getchar();
    while (isdigit(ch)) x = x *  + ch - '', ch = getchar(); return x * zf;
}
void write(ll y){if(y<)putchar('-'),y=-y;if(y>)write(y/);putchar(y%+);}
void writeln(const ll y){write(y);putchar('\n');}
int i,j,k,m,x,y,z,cnt,sz;ll n;
int sp[M],ss[M];bool pri[M];
int id1[M],id2[M],t,h[M];ll g[M],q[M];
#define inv2 500000004
void init(){
    sz=sqrt(n);
    for(rt i=;i<=sz;i++){
        if(!pri[i])ss[++cnt]=i,sp[cnt]=(sp[cnt-]+i)%p;
        for(rt j=;i*ss[j]<=sz&&j<=cnt;j++){
            pri[i*ss[j]]=;
            if(i%ss[j]==)break;
        }
    }
    for(ll i=;i<=n;){
        ll v=n/i;ll R=n/v;q[++t]=v;
        if(v<=sz)id1[v]=t;else id2[R]=t;
        v%=p;h[t]=v-;g[t]=v*(v+)%p*inv2%p-;
        i=R+;
    }
}
inline int id(const ll x){return x<=sz?id1[x]:id2[n/x];}
int S(ll x,int y){
    if(x<=||ss[y]>x)return ;
    ll ret=g[id(x)]-sp[y-]+y-;
    for(rt i=y;(ll)ss[i]*ss[i]<=x&&i<=cnt;i++)
    for(ll j=ss[i],e=;j*ss[i]<=x;j*=ss[i],e++)
    (ret+=(ll)(ss[i]^e)*S(x/j,i+)+(ss[i]^e+))%=p;

    return ret;
}
int main(){
    n=read();init();
    for(rt j=;j<=cnt;j++)
    for(rt i=;i<=t&&q[i]>=(ll)ss[j]*ss[j];i++){
        const int k=id(q[i]/ss[j]);
        (g[i]-=(ll)ss[j]*(g[k]-sp[j-]))%=p;
        (h[i]-=(h[k]-j+))%=p;
    }
    for(rt i=;i<=t;i++)(g[i]-=h[i])%=p;
    if(n==)cout<<;else cout<<(S(n,)++p)%p;
    return ;
}