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?

Input

The first line of input contains three integers, nk (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.

Output

For each question, output a single integer on a line by itself, the number of admissible integer temperatures between a and b degrees, inclusive.

Sample Input

Input
3 2 4
91 94
92 97
97 99
92 94
93 97
95 96
90 100
Output
3
3
0
4
Input
2 1 1
1 1
200000 200000
90 100
Output
0

Hint

In the first test case, Karen knows 3 recipes.

  1. The first one recommends brewing the coffee between 91 and 94 degrees, inclusive.
  2. The second one recommends brewing the coffee between 92 and 97 degrees, inclusive.
  3. 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.

  1. The first one, "wikiHow to make Cold Brew Coffee", recommends brewing the coffee at exactly 1 degree.
  2. 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.

  先给出n个区间,然后q次查询,每次查询 问在查询区间中有多个数 在开始给给定的n个区间中出现次数大于等于k(题意大致是这样)。

  前缀和。

对于开始给定的区间,利用括号匹配的思想,区间的左端点看作是左括号,右端点看作是右括号。然后从前向后遍历一遍,左括号每遇到一个右括号,就同时消去这一对括号,判断该数带有几个左括号

即为在几个区间出现过。前缀和完毕。

  

  

 #include<cstdio>
#include<cstdlib>
#include<cstring>
#include<string>
#include<cmath>
#include<algorithm>
#include<queue>
#include<stack>
#include<deque>
#include<map>
#include<iostream>
using namespace std;
typedef long long LL;
const double pi=acos(-1.0);
const double e=exp();
const int N = ; struct con{
int x;
int left,right;
}con[*N]; int num[*N]; int main()
{
int n,k,q;
int i,p,j;
int a,b;
scanf("%d%d%d",&n,&k,&q);
for(i=;i<=n;i++)
{
scanf("%d%d",&a,&b);
con[a].left++;
con[b+].right++;
}
a=b=;
for(i=;i<=*N;i++)
{
a+=con[i].left;
a-=con[i].right;
if(a>=k)
num[i]=;
num[i]+=num[i-];
}
// for(i=1;i<=100;i++)
// {
// printf("%d ",num[i]);
// if(i%10==0)
// putchar('\n');
// }
for(i=;i<=q;i++)
{
scanf("%d%d",&a,&b);
printf("%d\n",num[b]-num[a-]);
} return ;
}

CodeForces 816B 前缀和的更多相关文章

  1. CodeForces 816B Karen and Coffee(前缀和,大量查询)

    CodeForces 816B Karen and Coffee(前缀和,大量查询) Description Karen, a coffee aficionado, wants to know the ...

  2. Karen and Coffee CodeForces - 816B (差分数组+预处理前缀和)

    To stay woke and attentive during classes, Karen needs some coffee! Karen, a coffee aficionado, want ...

  3. Codeforces Round #427 (Div. 2) Problem C Star sky (Codeforces 835C) - 前缀和

    The Cartesian coordinate system is set in the sky. There you can see n stars, the i-th has coordinat ...

  4. codeforces 816B.Karen and Coffee 解题报告

    题目链接:http://codeforces.com/contest/816/problem/B 题目意思:给出 n 个recipes,第 i 个(1<= i <=n)recipes 表明 ...

  5. Codeforces 612D 前缀和处理区间问题

    传送门:http://codeforces.com/problemset/problem/612/D (转载请注明出处谢谢) 题意: 给出数字n和k,n表示接下来将输入n个在x轴上的闭区间[li,ri ...

  6. 【codeforces 816B】Karen and Coffee

    [题目链接]:http://codeforces.com/contest/816/problem/B [题意] 给你很多个区间[l,r]; 1<=l<=r<=2e5 一个数字如果被k ...

  7. codeforces 816B Karen and Coffee (差分思想)

    题目链接 816B Karen and Coffee 题目分析 题意:有个人在学泡咖啡,因此看了很多关于泡咖啡温度的书,得到了n种推荐的泡咖啡温度范围[L1,R1] ,此人将有k种做法推荐的温度记为可 ...

  8. Dasha and Photos CodeForces - 761F (前缀优化)

    大意: 给定n*m初始字符矩阵, 有k个新矩阵, 第$i$个矩阵是由初始矩阵区间赋值得到的, 求选择一个新矩阵, 使得其余新矩阵到它距离和最小. 字符集比较小, 可以考虑每次区间覆盖对每个字符的贡献. ...

  9. Codeforces 975 前缀和二分算存活人数 思维离直线速度相同判平行

    A /* Huyyt */ #include <bits/stdc++.h> using namespace std; typedef long long ll; ]; ]; map< ...

随机推荐

  1. scrapy-继承默认的user-agent 中间件

    class MyUserAgentMiddleware(UserAgentMiddleware): def __init__(self, user_agent): self.user_agent = ...

  2. day1 学习历程

    day1 我是一个在校大三学生,一个依然迷茫不知前景的大学混子= =,可以这么说吧 大学混子 真正开始决定好好学习大概在去年的12月份 那时经老师的提醒 开始正式接触软件开发 于是 从头开始学习语言 ...

  3. 2018 桂林ccpc现场赛 总结

    Day 0 5个小时的火车,坐的昏昏欲睡.桂林站出来没有地铁,而是出租车排成长队依次上车,也算是某种意义上的地铁吧.到了酒店才发现学校那边又给我们换了,又拖着行李找新的酒店,途中路过一家餐馆,所有人都 ...

  4. JVM工作原理 - 内存空间

    大多数 JVM 将内存区域划分为 Method Area(Non-Heap)(方法区) ,Heap(堆) , Program Counter Register(程序计数器) ,   VM Stack( ...

  5. Android Emoji兼容包使用详解

    Emoji兼容性 我们经常会遇到这样的问题: 给朋友发的emoji表情, 在自己手机上展示是正常的, 但是到朋友手机上, 却没有展示出来, 或者展示出来了, 但是也跟自己手机上展示的不一样. 所以,  ...

  6. [CQOI2016] 手机号码 (数位dp)

    link $solution:$ $10^{10} \leq L \leq R < 10^{11}$这个数据范围很容易想到数位$dp$. 依照题意模拟即可. #include<iostre ...

  7. SenseTime Ace Coder Challenge 暨 商汤在线编程挑战赛 A. 地铁站

    //其实比赛的时候就想到这方法了,但看到数据太吓人,就没写//看着标程,实际上就是这方法,太坑爹…… /* 假设值为k,对于图中任意两点,圆1半径k/t1,圆2半径k/t2 圆1与圆2的交集为可以设置 ...

  8. mysql 中的共享锁和排他锁

    共享锁(share lock) 共享锁又称读锁,是读取操作创建的锁.其他用户可以并发读取数据,但任何事务都不能对数据进行修改(获取数据上的排他锁),直到已释放所有共享锁. 如果事务T对数据A加上共享锁 ...

  9. web项目中配置文件的加载顺序

    当一个项目启动时,首先是web.xml: 这里面的配置: 为什么要在web.xml中配置struts的过滤器? 因为一个web项目运行的时需要加载的,或者默认的部分配置都会在web.xml中配置,中间 ...

  10. C语言复习---判断素数

    一般 int main01() { ; scanf("%d", &a); n_sqrt = sqrt(a); ; i <= n_sqrt; i++) ) { flag ...