原题传送门

817，我突然想到了某8位质数

这题珂以说是珂朵莉树的模板

三个操作都肥肠简单，前两个区间赋值，第三个区间0变1,1变0

每次输出从头开始扫描就行（我忘了珂朵莉树的性质，竟然还动态维护最左边0的位置）

#include <bits/stdc++.h>

#define getchar nc

#define ll long long

#define IT set<node>::iterator

using namespace std;

inline char nc(){

    static char buf[100000],*p1=buf,*p2=buf;

    return p1==p2&&(p2=(p1=buf)+fread(buf,1,100000,stdin),p1==p2)?EOF:*p1++;

}

inline ll read()

{

    register ll x=0,f=1;register 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;

}

inline void write(register ll x)

{

    if(!x)putchar('0');if(x<0)x=-x,putchar('-');

    static int sta[25];register int tot=0;

    while(x)sta[tot++]=x%10,x/=10;

    while(tot)putchar(sta[--tot]+48);

}

inline ll Min(register ll a,register ll b)

{

    return a<b?a:b;

}

struct node

{

   	ll l,r;

    mutable bool v;

    node(ll L, ll R=-1, bool V=0):l(L), r(R), v(V) {}

    bool operator<(const node& o) const

    {

        return l < o.l;

    }

};

set<node> s;

ll ans=1;

inline IT split(register ll pos)

{

    IT it = s.lower_bound(node(pos));

    if (it != s.end() && it->l == pos)

        return it;

    --it;

    ll L = it->l, R = it->r;

    bool V = it->v;

    s.erase(it);

    s.insert(node(L, pos-1, V));

    return s.insert(node(pos, R, V)).first;

}

inline void findnext()

{

    IT it = split(ans);

    for(;;++it)

        if(!it->v)

        {

            ans=it->l;

            return;

        }

}

inline void assign_val(register ll l,register ll r,register bool val)

{

    IT itr = split(r+1),itl = split(l);

    s.erase(itl, itr);

    s.insert(node(l, r, val));

    if(val&&l<=ans&&ans<=r)

        findnext();

    else if(!val)

        ans=Min(ans,l);

}

inline void reverse(register ll l,register ll r)

{

    IT itr = split(r+1),itl = split(l);

    bool f=true;

    if(l<=ans&&ans<=r)

    {

        for(; itl != itr; ++itl)

        {

            itl->v^=1;

            if(f&&!itl->v)

            {

                f=false;

                ans=itl->l;

            }

        }

        if(f)

            findnext();

    }

    else

        for(; itl != itr; ++itl)

        {

            itl->v^=1;

            if(f&&!itl->v)

            {

                f=false;

                ans=Min(ans,itl->l);

            }

        }

}

int main()

{

    int q=read();

    s.insert(node(1,1e19,0));

    while(q--)

    {

        int opt=read();

        ll l=read(),r=read();

        if(opt==1)

            assign_val(l,r,1);

        else if(opt==2)

            assign_val(l,r,0);

        else

            reverse(l,r);

        write(ans),puts("");

    }

    return 0;

}