Y is always discrete and can't understand, how to solve it? Answer: read it several times

y Always discrete and can't understand , How to solve ？ Answer: read it several times

I saw y Total discrete and , It was supposed to collapse , But the barrage track ： Look at it several times. . Do not deceive me , Here's a record , Also wrote more specific notes .
Challenge time , Suddenly I found that my hand speed increased a lot , The program is simple and happy

Interval and

Suppose there is an infinite number axis , The number on each coordinate on the number axis is 00.

Now? , Let's start with nn operations , Each operation will a certain position xx Add... To the number on cc.

Next , Conduct mm Time to ask , Each query contains two integers ll and rr, You need to find the interval [l,r][l,r] The sum of all numbers between .

Input format

The first line contains two integers nn and mm.

Next nn That's ok , Each line contains two integers xx and cc.

Next mm That's ok , Each line contains two integers ll and rr.

Output format

common mm That's ok , Each line outputs a number in the interval and .

Data range

−109≤x≤109−109≤x≤109,
1≤n,m≤1051≤n,m≤105,
−109≤l≤r≤109−109≤l≤r≤109,
−10000≤c≤10000

#include <iostream>
#include<vector>
#include<algorithm>
using namespace std;
typedef pair<int,int> pll;
vector<pll> adds,query; 
vector<int> alls; // Store all the coordinates that need to be operated in 
const int N = 300010;
int a[N],s[N];
// In fact, the discretization of large coordinates , Is the element value of the array, that is alls[i] It represents the value of the big target , Using discretization, the large scale can be corresponding to the small scale i, We use small coordinates to process . I think I've seen it several times , That's pretty clear .
// Use dichotomy to determine the number of small marks corresponding to the original big mark 
int find(int x){
    
    int l = 0, r = alls.size();
    while(l < r){
    
        int mid = l + r >> 1;
        if(alls[mid] >= x) r = mid;
        else l = mid + 1;
    } 
    return r + 1;
}
int main(){
    
    int n,m; cin >> n >> m;
    for(int i = 0; i < n; i++) {
    
        int x, c;
        cin >> x >> c;
        alls.push_back(x);
        adds.push_back({
    x,c});
    }

    for(int i = 0; i < m; i++){
    
        int l,r;
        cin >> l >> r;
        query.push_back({
    l,r});
        alls.push_back(l);
        alls.push_back(r);
    }

    sort(alls.begin(),alls.end());
    alls.erase(unique(alls.begin(),alls.end()),alls.end());  // duplicate removal 

    for(auto item: adds){
      //  The first step in implementing the requirements of the topic 
        int x = find(item.first);  // Find the small coordinate corresponding to the large coordinate 
        a[x] += item.second;
    }
	// Prefixes and preprocessing 
    for(int i = 1; i <= alls.size(); i++) s[i] = s[i-1] + a[i];
    // Perform the second step of the topic 
    for(auto item : query){
     
        int l = find(item.first), r = find(item.second);
        cout << s[r] - s[l-1] << endl;
    }
}