Codeforces Round #419 (Div. 2) B. Karen and Coffee(经典前缀和)
Karen, a coffee aficionado, wants to know the optimal temperature for brewing the perfect cup of coffee. Indeed, she has spent some time reading several recipe books, including the universally acclaimed "The Art of the Covfefe".
She knows n coffee recipes. The i-th recipe suggests that coffee should be brewed between li and ri degrees, inclusive, to achieve the optimal taste.
Karen thinks that a temperature is admissible if at least k recipes recommend it.
Karen has a rather fickle mind, and so she asks q questions. In each question, given that she only wants to prepare coffee with a temperature between a and b, inclusive, can you tell her how many admissible integer temperatures fall within the range?
The first line of input contains three integers, n, k (1 ≤ k ≤ n ≤ 200000), and q (1 ≤ q ≤ 200000), the number of recipes, the minimum number of recipes a certain temperature must be recommended by to be admissible, and the number of questions Karen has, respectively.
The next n lines describe the recipes. Specifically, the i-th line among these contains two integers li and ri (1 ≤ li ≤ ri ≤ 200000), describing that the i-th recipe suggests that the coffee be brewed between li and ri degrees, inclusive.
The next q lines describe the questions. Each of these lines contains a and b, (1 ≤ a ≤ b ≤ 200000), describing that she wants to know the number of admissible integer temperatures between a and b degrees, inclusive.
For each question, output a single integer on a line by itself, the number of admissible integer temperatures between a and b degrees, inclusive.
3 2 4
91 94
92 97
97 99
92 94
93 97
95 96
90 100
3
3
0
4
2 1 1
1 1
200000 200000
90 100
0
给出n个区间,然后进行q次区间查询,问每次查询区间中的数至少在k个区间中出现的个数。
思路:
前缀和的经典应用。
也就是++cnt【l】和--cnt【r+1】,很妙。
#include<iostream>
#include<algorithm>
#include<cstring>
#include<cstdio>
#include<sstream>
#include<vector>
#include<stack>
#include<queue>
#include<cmath>
#include<map>
#include<set>
using namespace std;
typedef long long ll;
typedef pair<int,int> pll;
const int INF = 0x3f3f3f3f;
const int maxn = +; int n, k, q;
int cnt[maxn];
int sum[maxn]; int main()
{
//freopen("in.txt","r",stdin);
while(~scanf("%d%d%d",&n, &k, &q))
{
memset(cnt,,sizeof(cnt));
for(int i=;i<n;i++)
{
int l, r;
scanf("%d%d",&l,&r);
++cnt[l];
--cnt[r+];
} memset(sum,,sizeof(sum));
for(int i=;i<=maxn;i++)
{
cnt[i]+=cnt[i-];
if(cnt[i]>=k) sum[i]=;
sum[i]+=sum[i-];
} while(q--)
{
int l, r;
scanf("%d%d",&l, &r);
printf("%d\n",sum[r]-sum[l-]);
}
}
return ;
}
Codeforces Round #419 (Div. 2) B. Karen and Coffee(经典前缀和)的更多相关文章
- Codeforces Round #419 (Div. 2) B. Karen and Coffee
To stay woke and attentive during classes, Karen needs some coffee! Karen, a coffee aficionado, want ...
- Codeforces Round #419 (Div. 2) C. Karen and Game
C. Karen and Game time limit per test 2 seconds memory limit per test 512 megabytes input standard i ...
- Codeforces Round #419 (Div. 2) E. Karen and Supermarket(树形dp)
http://codeforces.com/contest/816/problem/E 题意: 去超市买东西,共有m块钱,每件商品有优惠卷可用,前提是xi商品的优惠券被用.问最多能买多少件商品? 思路 ...
- Codeforces Round #419 (Div. 2) A. Karen and Morning(模拟)
http://codeforces.com/contest/816/problem/A 题意: 给出一个时间,问最少过多少时间后是回文串. 思路: 模拟,先把小时的逆串计算出来: ① 如果逆串=分钟, ...
- Codeforces Round #419 (Div. 1) C. Karen and Supermarket 树形DP
C. Karen and Supermarket On the way home, Karen decided to stop by the supermarket to buy some g ...
- 【找规律】【递推】【二项式定理】Codeforces Round #419 (Div. 1) B. Karen and Test
打个表出来看看,其实很明显. 推荐打这俩组 11 1 10 100 1000 10000 100000 1000000 10000000 100000000 1000000000 1000000000 ...
- 【贪心】 Codeforces Round #419 (Div. 1) A. Karen and Game
容易发现,删除的顺序不影响答案. 所以可以随便删. 如果行数大于列数,就先删列:否则先删行. #include<cstdio> #include<algorithm> usin ...
- Codeforces Round #419 (Div. 2) A-E
上紫啦! E题1:59压哨提交成功翻盘 (1:00就做完了调了一个小时,还好意思说出来? (逃)) 题面太长就不复制了,但是配图很可爱所以要贴过来 九条可怜酱好可爱呀 A - Karen and Mo ...
- Codeforces Round #419 (Div. 2)
1.题目A:Karen and Morning 题意: 给出hh:mm格式的时间,问至少经过多少分钟后,该时刻为回文字符串? 思路: 简单模拟,从当前时刻开始,如果hh的回文rh等于mm则停止累计.否 ...
随机推荐
- UUID的定义以及作用
UUID含义是通用唯一识别码 (Universally Unique Identifier),这 是一个软件建构的标准,也是被开源软件基金会 (Open Software Foundation, OS ...
- poj3764 The XOR Longest Path【dfs】【Trie树】
The xor-longest Path Time Limit: 2000MS Memory Limit: 65536K Total Submissions: 10038 Accepted: ...
- A Benchmark Comparsion of Monocular Visual-Inertial Odometry Algorithms for Flying Robots论文笔记
摘要: 本文主要比较单目VIO的算法在飞行机器人上运行的性能,测试使用统一数据集为EuRoC.其中评价指标为:姿态估计精度.每帧处理时间以及CPU和内存负载使用率,同时还有RMSE(运行轨迹与真实轨迹 ...
- StartUML-类图
- Maven 的聚合(多模块)和 Parent 继承
2017年06月26日 21:16:57 Maven 的聚合(多模块)和 Parent 继承 - 偶尔记一下 - CSDN博客 https://blog.csdn.net/isea533/articl ...
- Romantic---hdu2669(扩展欧几里德模板)
题目链接:http://acm.hdu.edu.cn/showproblem.php?pid=2669 详解:扩展欧几里德 #include <iostream> #include < ...
- Mirror--镜像相关错误
场景: 同一局域网下搭建了镜像的服务器上报以下错误日志: An error occurred in a Service Broker/Database Mirroring transport conn ...
- Openstack(二)基本环境准备--网络、时间、yum源等
2.1服务器版本安装 2.1.1服务器使用:centos7.4 + vm12 2.1.2重命名网卡: 传递内核参数 net.ifnames=0 biosdevname=0,以更改网卡名称为eth0,e ...
- sdut AOE网上的关键路径(spfa+前向星)
http://acm.sdut.edu.cn/sdutoj/showproblem.php?pid=2498&cid=1304 题目描述 一个无环的有向图称为无环图(Directed Acyc ...
- django开发项目的部署nginx
Django 部署(Nginx) 本文主要讲解 nginx + uwsgi socket 的方式来部署 Django,比 Apache mod_wsgi 要复杂一些,但这是目前主流的方法. 1. 运行 ...