【NOI2017】整數

題目簡述：
一開始你有一個數x=0$x=0$
n(n≤106)$n(n\le {10}^6)$次操作:
1.給x$x$加上a×2k$a\times 2^k$ (a≤109,k≤3×107)$(a\le {10}^9,k\le 3\times {10}^7)$
2.查詢x$x$的二進位制表達表示2b(b≤3×107)$2^b(b\le 3\times {10}^7)$的那位的值
根據最基本的均攤分析，我們知道只支援加的二進位制計數器均攤複雜度是O(1)$O(1)$的
所以如果本題只支援加法的話可以利用unsignedint$unsignedint$壓位
複雜度可以達到O(30n32)=O(n)$O(\frac{30n}{32})=O(n)$
但是可能產生減法怎麼辦？
大力維護一個加法的結果
再維護一個減法的結果
查詢怎麼辦？
大力

#include<iostream>
#include<cstdio>
#include<cstring>
#include<algorithm>
#include<set>
using namespace std;
const int MAXN=1000010;
typedef unsigned long
 
 long lsq;
lsq Plus[MAXN],Minus[MAXN];
set<int>st;
set<int>::iterator it;
int n,t1,t2,t3,opt;
int main()
{
    scanf("%d%d%d%d",&n,&t1,&t2,&t3);
    while(n--)
    {
        scanf("%d",&opt);
        if(opt==1)
        {
            int a,b;
            scanf("%d%d",&a,&b);
            int
 
 pos1=b/64,pos2=b%64;
            if(a>0)
            {
                lsq temp=Plus[pos1];
                Plus[pos1]+=(lsq)(a<<pos2);
                lsq up=(lsq)(a>>(31-pos2));
                up>>=31;
                up>>=2;
                up+=(Plus[pos1]<temp);
                if(Plus[pos1]^Minus[pos1])
                    st.insert(pos1);
                else if(st.count(pos1))st.erase(pos1);
                ++pos1;
                while(up)
                {
                    temp=Plus[pos1];
                    Plus[pos1]+=up;
                    up=(Plus[pos1]<temp);
                    if(Plus[pos1]^Minus[pos1])
                        st.insert(pos1);
                    else if(st.count(pos1))st.erase(pos1);
                    ++pos1;
                }
            }
            else if(a<0)
            {
                a=-a;
                lsq temp=Minus[pos1];
                Minus[pos1]+=(lsq)(a<<pos2);
                lsq up=(lsq)(a>>(31-pos2));
                up>>=31;
                up>>=2;
                up+=(Minus[pos1]<temp);
                if(Plus[pos1]^Minus[pos1])
                    st.insert(pos1);
                else if(st.count(pos1))st.erase(pos1);
                ++pos1;
                while(up)
                {
                    temp=Minus[pos1];
                    Minus[pos1]+=up;
                    up=(Minus[pos1]<temp);
                    if(Plus[pos1]^Minus[pos1])
                        st.insert(pos1);
                    else if(st.count(pos1))st.erase(pos1);
                    ++pos1;
                }
            }
        }
        else
        {
            int b;
            scanf("%d",&b);
            int pos1=b/64,pos2=b%64;
            int ans=(lsq)((Plus[pos1]>>pos2)^(Minus[pos1]>>pos2))&1;
            lsq valp=(lsq)Plus[pos1]%(1<<pos2),valm=(lsq)Minus[pos1]%(1<<pos2);
            if(valp<valm)printf("%d\n",(ans^1));
            else if(valp>valm||(!st.size())||pos1<=*(st.begin()))
                printf("%d\n",ans);
            else
            {
                it=st.lower_bound(pos1);
                --it;
                if(Plus[*it]>Minus[*it])printf("%d\n",ans);
                else printf("%d\n",(ans^1));
            }
        }
    }
}