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
- Python 中的线程-进程2
原文:https://www.cnblogs.com/i-honey/p/7823587.html Python中实现多线程需要使用到 threading 库,其中每一个 Thread类 的实例控制一 ...
- Nodejs Web模块( readFile 根据请求跳转到响应html )
index.js 根据请求的路径pathname,返回响应的页面. var http = require('http'); var fs = require('fs'); var url = requ ...
- c# 对数据库的操作
1.首先需要引用 using System.Data.SqlClient; 2.创建连接 SqlConnection connection = new SqlConnection(); connect ...
- 系统事件管理(Events) ---- HTML5+
模块:events Events模块管理客户端事件,包括系统事件,如扩展API加载完毕.程序前后台切换等. 比如说:网络的链接的和断开这种事件,系统从前台走到后台这种事件: 不包括:点击和滑动页面事件 ...
- 110道python题+理解(不断更新)
此篇题目在网上已经广为流传,但好多都不做解释,所以我想着自己一道一道的做一遍,并将相关涉及的做个补充,个人知识毕竟片面,有不足的地方还请大家多多指正 一.请用一行代码实现1-100之和 >> ...
- SQL中 decode()函数简介(转载)
今天看别人的SQL时看这里面还有decode()函数,以前从来没接触到,上网查了一下,还挺好用的一个函数,写下来希望对朋友们有帮助哈! decode()函数简介: 主要作用:将查询结果翻译成其他值(即 ...
- Oracle管理监控之使用utl_mail自动邮件报警配置
--代发邮件存储过程源码如下: CREATE OR REPLACE PROCEDURE send_mail(p_recipient VARCHAR2, -- 邮件接收人 ...
- 数字签名中公钥和私钥是什么?对称加密与非对称加密,以及RSA的原理
http://baijiahao.baidu.com/s?id=1581684919791448393&wfr=spider&for=pc https://blog.csdn.net/ ...
- Django - 路由层(URLconf)
一.django 静态文件配置 /mysite1/settings.py STATIC_URL = '/static/' STATICFILES_DIRS = [ os.path.join(BASE_ ...
- ConcurrentHashMap实现解析
ConcurrentHashMap是线程安全的HashMap的实现,具有更加高效的并发性.与HashTable不同,ConcurrentHashMap运用锁分离技术,尽量减小写操作时加锁的粒度,即在写 ...