luogu_P3674 小清新人渣的本愿

传送门

Solution

莫队，用bitset来存储出现的数

如果是和或者差，直接通过左移右移就可以实现判断

对于积的询问，暴力判就行了，因数只要枚举(sqrt n)个

总复杂度是(O(n^2/32))，反正(3s)是可以过的咯

Code 

#include<bits/stdc++.h>
#define ll long long
#define max(a,b) ((a)>(b)?(a):(b))
#define min(a,b) ((a)<(b)?(a):(b))
inline int read()
{
    int x=0,f=1;char ch=getchar();
    while(ch<'0'||ch>'9'){if(ch=='-')f=-1;ch=getchar();}
    while(ch>='0'&&ch<='9'){x=(x<<3)+(x<<1)+ch-'0';ch=getchar();}
    return x*f;
}
#define MN 100005
#define N MN
int n,m,a[MN],T;
bool Ans[MN];
std::bitset<N> now,fnow;
struct ques{
    int l,r,opt,x,id,pl;
    bool operator <(const ques&o)const{return (pl^o.pl)?(pl<o.pl):(r<o.r);}
}q[MN];
int num[MN];
int main()
{
    //freopen("testdata.in","r",stdin);
    //freopen("testdata.out","w",stdout);
    n=read();m=read();
    register int i;T=ceil(sqrt((double)n));
    for(i=1;i<=n;++i) a[i]=read();
    now.reset();fnow.reset();
    for(i=1;i<=m;++i)
    {
        q[i].opt=read(),q[i].l=read(),q[i].r=read();
        q[i].x=read(),q[i].pl=(q[i].l-1)/T+1;q[i].id=i;
    }
    std::sort(q+1,q+m+1);
    register int l=1,r=0,j;
    for(i=1;i<=m;++i)
    {
        //printf("l=%d r=%d
",q[i].l,q[i].r);
        for(;r<q[i].r;++r) if(!num[a[r+1]]++) now[a[r+1]]=1,fnow[N-a[r+1]]=1;
        for(;l>q[i].l;--l) if(!num[a[l-1]]++) now[a[l-1]]=1,fnow[N-a[l-1]]=1;
        for(;r>q[i].r;--r) if(!(--num[a[r]])) now[a[r]]=0,fnow[N-a[r]]=0;
        for(;l<q[i].l;++l) if(!(--num[a[l]])) now[a[l]]=0,fnow[N-a[l]]=0;
        //std::cout<<now<<std::endl<<fnow<<std::endl;
        if(q[i].opt==1) Ans[q[i].id]=(now&(now<<q[i].x)).any();
        else if(q[i].opt==2) Ans[q[i].id]=((now<<(N-q[i].x))&fnow).any();
        else
        {
            for(j=1;j*j<=q[i].x;j++)
                if(q[i].x%j==0) if(now[j]&&now[q[i].x/j]){Ans[q[i].id]=1;break;}
        }
    }
    for(i=1;i<=m;++i) puts(Ans[i]?"hana":"bi");
    return 0;
}

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