CF1097F Alex and a TV Show
题目
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;
}
