题目描述

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样例

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算法1

(离散化) $O(n)$

blablabla

时间复杂度

参考文献

C++ 代码

#include<iostream>
#include<vector>
#include<algorithm>
using namespace std;

const int N = 3e5 + 10;
int a[N],s[N];
int n,m;

typedef pair<int,int> PII;

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

int find(int x)
{
    int l = 0,r = alls.size()-1;
    while(l < r)
    {
        int mid = (l + r)/2;
        if(alls[mid] >= x) r = mid;
        else l = mid + 1;
    }
    return l + 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 l,r;
        cin >> l >> r;
        query.push_back({l,r});
        alls.push_back(l);
        alls.push_back(r);
    }

    sort(alls.begin(),alls.end());    //排序
    alls.erase(unique(alls.begin(),alls.end()),alls.end());   //去重

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

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

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

算法2

(暴力枚举) $O(n^2)$

blablabla

时间复杂度

参考文献

C++ 代码

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