把所有坐标都放入alls中离散化，然后在正常操作即可

#include<bits/stdc++.h>
using namespace std;
typedef pair<int,int> P;
const int N=3e5+10;

int n,m;
int a[N],s[N];

vector<int> alls;
vector<P> add,query;

int find(int x){
    int l=0,r=alls.size()-1;
    while(l<r){
        int mid=l+r>>1;
        if(alls[mid]>=x) r=mid;
        else l=mid+1;
    }
    return r+1;
}

int main(){
    cin>>n>>m;
    for(int i=0;i<n;i++){
        int x,c;
        cin>>x>>c;

        add.push_back({x,c});
        alls.push_back(x);
    }

    for(int i=0;i<m;i++){
        int a,b;
        cin>>a>>b;

        query.push_back({a,b});
        alls.push_back(a);
        alls.push_back(b);
    }

    sort(alls.begin(),alls.end());
    alls.erase(unique(alls.begin(),alls.end()),alls.end());


    for(auto item:add){
        int x=find(item.first);
        a[x]+=item.second;
    }

    for(int i=1;i<=alls.size();i++) s[i]=s[i-1]+a[i];

    for(auto item:query){
        int l=find(item.first),r=find(item.second);
        cout<<s[r]-s[l-1]<<endl;
    }
    return 0;
}