# 10115 "through 4.1 cases a 3" tree outside the school

1537: [Example 3] tree outside the school

Time limit: 1000 ms Memory Limit: 524288 KB
Submissions: 497 number by: 216 

Description [title]

From the original title: Vijos P1448

There are a lot of trees outside the school, the school decided at some point in a certain period of planted tree with guarantee any one time two of the same kind of tree does not appear, the existing two actions:

K = . 1 K =. 1, the read  L , r L, r represents  L L to  r planted tree with between r, the type of each operation are different kinds of trees;

K = 2 K = 2, the read  L , r L, r represents interrogation  L L the  r number r between the trees.

Note: Each location can be repeated to plant trees.

[Enter]

The first line  n- , m n-, m represents a road length as  n- n-, a total of  m m operations;

Subsequently  m m behavior  m m operations.

[Output]

For each of the  K = 2 K = 2 output an answer.

[Sample input]

5 4
1 1 3
2 2 5
1 2 4
2 3 5

[Sample Output]

1
2

【prompt】

Data range and tips:

For 20% of the data, . 1 ≤ n- , m ≤ 100 1 ≦ n, m≤100;

% To 60% of the data, . 1 ≤ n- ≤ 10 . 3 , . 1 ≤ m ≤ . 5 × 10 . 4 1≤n≤103,1≤m≤5 × 104;

% For 100% of the data, . 1 ≤ n- , m ≤ . 5 × 10 . 4 1 ≦ n, m≤5 × 104, to ensure  L , R & lt > 0 L, R & lt> 0.

【source】

no

 

#include<bits/stdc++.h>
using namespace std;
int c1[50005],c2[50005];
int low(int x){return x&(-x);}
int find(int l,int r){
    int ans=0;
    for(int i=r;i;i-=low(i))ans+=c1[i];
    for(int i=l-1;i;i-=low(i))ans-=c2[i];
    return ans;
}
int main(){
    int n,m;
    scanf("%d%d",&n,&m);
    for(int i=1;i<=m;i++){
        int k,l,r;
        scanf("%d%d%d",&k,&l,&r);
        if(k==1){
            for(int i=l;i<=n+1;i+=low(i))c1[i]++;
            for(int i=r;i<=n+1;i+=low(i))c2[i]++;
        }
        else printf("%d\n",find(l,r));
    }
}