2019.9.4 simple question

Yes you read that right it really be called simple question

Do not believe you

Title Description

An array of n elements, each element initially are 0. M has instructions, which allow for a continuous sequence or digit reversal becomes 0 --0 variants 1,1 (operation 1), or an element of the challenge value (operation 2). For example, when n = 20, the 10 following instructions:

Input Format

The first line contains two integers n, m, and the length of the bar represents the number of instructions of the array; the m rows, each row of the first number and t represents the type of the operation:

When t = 1, then the next two numbers L, R, representing the interval [L, R] is inverted for each number; if t = 2, only one number is next i, subscript represents inquiry.

Output Format

Each operation of the second output line (i.e. a non-zero), 2 represents the operation of each answer.

Sample input and output

Input # 1copy

20 10
1 1 10
2 6
2 12
1 5 12
2 6
2 15
1 6 16
1 11 17
2 12
2 6

Output # 1copy

1
0
0
0
1
1

Description / Tips

For 50% of the data,. 1 ≤ n-≤  10 ^. 3 . 1 0 . 3,. 1 ≤ m ≤  10 ^. 4 . 1 0 . 4; 100% of the data,. 1 ≤ n-≤  10 ^. 5 . 1 0 . 5,. 1 ≤ m ≤. 5 ×  10 ^. 5 . 1 0 . 5, to ensure that L ≤ R.

Title source CQOI2006 / Luo Gu P5057

---

And the principle that a tree outside the school but this problem is to modify a single point of inquiry interval

Also considered for each point by the length of the sequence in parentheses is the method of section 1 can

Note 1 modified once becomes twice becomes 0

The Code

#include<iostream>
#include<cstdio>
#include<cstring>
#define int long long
#define lowbit(x) x&(-x)
using namespace std;
int n,m,t,ll,rr,xx,l[1000050],r[1000050];
void updatal(int i,int x)
{
    while(i<=n)
    {
        l[i]+=x;
        i+=lowbit(i);
    }
}
void updatar(int i,int x)
{
    while(i<=n)
    {
        r[i]+=x;
        i+=lowbit(i);
    }
}
int queryl(int i)
{
    int res=0;
    while(i>=1)
    {
        res+=l[i];
        i-=lowbit(i);
    }
    return res;
}
int queryr(int i)
{
    int res=0;
    while(i>=1)
    {
        res+=r[i];
        i-=lowbit(i);
    }
    return res;
}
signed main()
{
    scanf("%lld%lld",&n,&m);
    for(int i=1;i<=m;i++)
    {
        scanf("%lld",&t);
        if(t==1)
        {
            scanf("%lld%lld",&ll,&rr);
            updatal(ll,1);
            updatar(rr,1);
        }
        if(t==2)
        {
            scanf("%lld",&xx);
            printf("%lld\n",(queryl(xx)-queryr(xx-1))%2);
        }
    }
    return 0;
}