analysis
Respectively, and r is the number of maintenance l
Then the inquiry interval [L, R]
After l tree array R less than or equal to the number of minus L r is less than the array tree
Code
#include<bits/stdc++.h>
using namespace std;
int n;
struct BIT {
int d[100100];
inline int lb(int x){return x&(-x);}
inline void add(int x){while(x<=n)d[x]++,x+=lb(x);}
inline int que(int x){int res=0;while(x)res+=d[x],x-=lb(x);return res;}
};
BIT a,b;
int main(){
int i,j,k,x,y;
cin>>n;
for(i=1;i<=n;i++){
scanf("%d",&k);
if(k==1)scanf("%d%d",&x,&y),a.add(x),b.add(y);
else scanf("%d%d",&x,&y),printf("%d\n",a.que(y)-b.que(x-1));
}
return 0;
}