codeforces round #419 B. Karen and Coffee
To stay woke and attentive during classes, Karen needs some coffee!

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
In the first test case, Karen knows 3 recipes.
- The first one recommends brewing the coffee between 91 and 94 degrees, inclusive.
- The second one recommends brewing the coffee between 92 and 97 degrees, inclusive.
- The third one recommends brewing the coffee between 97 and 99 degrees, inclusive.
A temperature is admissible if at least 2 recipes recommend it.
She asks 4 questions.
In her first question, she wants to know the number of admissible integer temperatures between 92 and 94 degrees, inclusive. There are 3: 92, 93 and 94 degrees are all admissible.
In her second question, she wants to know the number of admissible integer temperatures between 93 and 97 degrees, inclusive. There are 3: 93, 94 and 97 degrees are all admissible.
In her third question, she wants to know the number of admissible integer temperatures between 95 and 96 degrees, inclusive. There are none.
In her final question, she wants to know the number of admissible integer temperatures between 90 and 100 degrees, inclusive. There are 4: 92, 93, 94 and 97 degrees are all admissible.
In the second test case, Karen knows 2 recipes.
- The first one, "wikiHow to make Cold Brew Coffee", recommends brewing the coffee at exactly 1 degree.
- The second one, "What good is coffee that isn't brewed at at least 36.3306 times the temperature of the surface of the sun?", recommends brewing the coffee at exactly 200000 degrees.
A temperature is admissible if at least 1 recipe recommends it.
In her first and only question, she wants to know the number of admissible integer temperatures that are actually reasonable. There are none.
题解:
开一个线段树区间修改输入的东西,然后从1扫到200000 如果该位>=k就把该位赋为1
询问就是区间查询
水题,乱搞即可
#include<iostream>
#include<cstdio>
#include<cstring>
#include<algorithm>
#define ls (node<<1)
#define rs (node<<1|1)
using namespace std;
const int N=;
int Tree[][N*],mark[][N*];
void updata(bool t,int node){
Tree[t][node]=Tree[t][ls]+Tree[t][rs];
}
void pushdown(bool t,int node)
{
int k=mark[t][node];
mark[t][ls]+=k;mark[t][rs]+=k;
Tree[t][ls]+=k;Tree[t][rs]+=k;
mark[t][node]=;
}
void add(bool t,int l,int r,int node,int sa,int se)
{
if(l>se || r<sa)return ;
if(sa<=l && r<=se)
{
Tree[t][node]++;
mark[t][node]++;
return ;
}
pushdown(t,node);
int mid=(l+r)>>;
add(t,l,mid,ls,sa,se);
add(t,mid+,r,rs,sa,se);
updata(t,node);
}
int getsum(bool t,int l,int r,int node,int sa,int se)
{
if(l>se || r<sa)return ;
if(sa<=l && r<=se)return Tree[t][node];
int mid=(l+r)>>;
pushdown(t,node);
updata(t,node);
return getsum(t,l,mid,ls,sa,se)+getsum(t,mid+,r,rs,sa,se);
}
int main()
{
int n,m,q,x,y,tmp;
scanf("%d%d%d",&n,&m,&q);
for(int i=;i<=n;i++)
{
scanf("%d%d",&x,&y);
add(,,,,x,y);
}
for(int i=;i<=;i++)
{
tmp=getsum(,,,,i,i);
if(tmp>=m)add(,,,,i,i);
}
for(int i=;i<=q;i++)
{
scanf("%d%d",&x,&y);
printf("%d\n",getsum(,,,,x,y));
}
return ;
}
codeforces round #419 B. Karen and Coffee的更多相关文章
- Codeforces Round #419 D. Karen and Test
Karen has just arrived at school, and she has a math test today! The test is about basic addition an ...
- codeforces round #419 E. Karen and Supermarket
On the way home, Karen decided to stop by the supermarket to buy some groceries. She needs to buy a ...
- codeforces round #419 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 A. Karen and Morning
Karen is getting ready for a new school day! It is currently hh:mm, given in a 24-hour format. As yo ...
- Codeforces Round #419 (Div. 2) B. Karen and Coffee(经典前缀和)
http://codeforces.com/contest/816/problem/B To stay woke and attentive during classes, Karen needs s ...
- 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)
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 ...
- Codeforces Round #419 A+B
A. Karen and Morning time limit per test 2 seconds memory limit per test 512 megabytes Karen is ...
- Codeforces Round #419 (Div. 2) A B C 暴力 区间更新技巧 +前缀和 模拟
A. Karen and Morning time limit per test 2 seconds memory limit per test 512 megabytes input standar ...
随机推荐
- 敏捷冲刺每日报告四(Java-Team)
第四天报告(10.28 周六) 团队:Java-Team 成员: 章辉宇(284) 吴政楠(286) 陈阳(PM:288) 韩华颂(142) 胡志权(143) github地址:https://gi ...
- Alpha集合
项目名称:城市安全风险管控系统 小组成员: 张梨贤.林静.周静平.黄腾飞 Alpha冲刺随笔 Alpha冲刺Day1 Alpha冲刺Day2 Alpha冲刺Day3 Alpha冲刺Day4 Alpha ...
- win7 Anaconda 安装 scrapy模块
之前用了很多方法,都安装不成功,今天终于成功了..说下方法.. anaconda的清华镜像:https://mirrors.tuna.tsinghua.edu.cn/anaconda/archive/ ...
- R语言-推荐系统
一.概述 目的:使用推荐系统可以给用户推荐更好的商品和服务,使得产品的利润更高 算法:协同过滤 协同过滤是推荐系统最常见的算法之一,算法适用用户过去的购买记录和偏好进行推荐 基于商品的协同过滤(IBC ...
- JAVA_SE基础——70.Math类
package cn.itcast.other; /* Math 数学类, 主要是提供了很多的数学公式. abs(double a) 获取绝对值 ceil(double a) 向上取整 ...
- Mybatis-select-返回值类型错误理解
Mybatis :Cause: java.lang.UnsupportedOperationException异常: 今天在写一个练手项目,作为初学Mybatis的小白,想着这里findByEmp_i ...
- PC或者手机弹出窗效果
http://layer.layui.com/ 这个网站提供弹窗,是在jq封装的,弹窗之后,背景页面还可以滑动. 这个里面的js可能也会包含css,这个css不能移动位置,否则会报错,还有谷歌浏览器在 ...
- 使用 BenchmarkDotnet 测试代码性能
先来点题外话,清明节前把工作辞了(去 tm 的垃圾团队,各种拉帮结派.勾心斗角).这次找工作就得慢慢找了,不能急了,希望能找到个好团队,好岗位吧.顺便这段时间也算是比较闲,也能学习一下和填掉手上的坑. ...
- Docker学习笔记 - Docker Compose
一.概念 Docker Compose 用于定义运行使用多个容器的应用,可以一条命令启动应用(多个容器). 使用Docker Compose 的步骤: 定义容器 Dockerfile 定义应用的各个服 ...
- 新概念英语(1-141)Sally's first train ride
Lesson 141 Sally's first train ride 萨莉第一交乘火车旅行 Listen to the tape then answer this question. Why was ...