思路:
维护两个树状数组分别记录L和R
当每询问一次的时候(坐标为x):
(1)在L中找到sum(x) 即左端点在区间[1,x]的个数n1
(2)在R中找到sum(x-1) 即右端点在区间[1,x-1]的个数n2
ans=(n1-n2)%2
代码如下:
#include<iostream>
#include<cstdio>
#include<algorithm>
#include<string>
#include<cstring>
#include<queue>
#include<stack>
#include<cmath>
#include<set>
#include<map>
using namespace std;
#define ll long long
#define lson l,m,rt<<1
#define rson m+1,r,rt<<1|1
typedef pair<int,int>P;
const int INF=0x3f3f3f3f;
const int N=500010;
int L[N],R[N];
int lowbit(int x){
return x&(-x);
}
int sum(int x,int c[]){
int res=0;
while(x>0){
res+=c[x];
x-=lowbit(x);
}
return res;
}
void add(int x,int c[]){
while(x<=N){
c[x]++;
x+=lowbit(x);
}
}
int main(){
int n,m,t,l,r,x;
scanf("%d%d",&n,&m);
while(m--){
scanf("%d",&t);
if(t==1){
scanf("%d%d",&l,&r);
add(l,L);
add(r,R);
}
else{
scanf("%d",&x);
int tmp=sum(x,L)-sum(x-1,R);
tmp%=2;
printf("%d\n",tmp);
}
}
}