CF1097F Alex and a TV Show

题目

CF1097F Alex and a TV Show

做法

奇偶性，考虑用\(bitset\)维护，\(Set[i][j]\)为第\(i\)个集合中\(j\)作为因子出现的次数

预处理\(yz[i]\)(\(i\)中的因子)

\(1\)：直接\(Set[x]=yz[y]\)

\(2\)：相加取奇偶相当于异或

\(3\)：相乘取奇偶相当于且

\(4\)：\(F(n)\)为\(n\)作为因子出现的次数，\(f(n)\)为\(n\)的出现次数
\[\begin{aligned} F(n)&=\sum\limits_{n|d}f(d)\\ f(n)&=\sum\limits_{n|d}{\mu(\lfloor\frac{d}{n}\rfloor)F(d)} \end{aligned}\]

My complete code

#include<cstdio>
#include<cstring>
#include<iostream>
#include<string>
#include<algorithm>
#include<bitset>
using namespace std;
typedef int LL;
inline LL Read(){
    LL x(0),f(1);char c=getchar();
    while(c<'0'||c>'9'){if(c=='-')f=-1;c=getchar();}
    while(c>='0'&&c<='9')x=(x<<3)+(x<<1)+c-'0',c=getchar();
    return x*f;
}
LL n,q;
bitset<7009> yz[7009],Mu[7009],Set[100009];
LL prime[7009],mu[7009];
bool visit[7009];
inline void First(LL N){
    mu[1]=1; LL tot(0);
    for(LL i=2;i<=N;++i){
        if(!visit[i]){
            mu[i]=-1;
            prime[++tot]=i;
        }
        for(LL j=1;j<=tot&&i*prime[j]<=N;++j){
            visit[prime[j]*i]=true;
            if((i%prime[j])==0)
                break;
            else
                mu[i*prime[j]]=-mu[i];
        }
    }
}
int main(){
    First(7000);
    for(LL i=1;i<=7000;++i)
        for(LL j=i;j<=7000;j+=i)
            yz[j][i]=1,Mu[i][j]=mu[j/i]!=0;
    n=Read(),q=Read();
    while(q--){
        LL op(Read()),x(Read()),y(Read());
        if(op==1){
            Set[x]=yz[y];
        }else if(op==2){
            LL z(Read());
            Set[x]=Set[y]^Set[z];
        }else if(op==3){
            LL z(Read());
            Set[x]=Set[y]&Set[z];
        }else
            printf("%d",(Set[x]&Mu[y]).count()&1);
    }
    return 0;
}