标签：200000 coffee 前缀 int Coffee admissible Karen between

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, 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.

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.

Examples

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

题目大意：
给出三个数n,k,q,接下来给出k个区间，区间内元素+1，然后给出q个查询，问查询区间内值大于等于k的元素有多少个

解题思路：
暴力会超时，用前缀和可以降低时间复杂度。

完整代码：
#include
using namespace std;
int main()
{
int n,k,q,a[200010]={0};
cin>>n>>k>>q;
while(n–){
int x,y;
scanf("%d%d",&x,&y);
a[x]++;
a[y+1]–;
}
for(int i=1;i<=200005;i++){ //伪前缀和
a[i]+=a[i-1];
}
for(int i=1;i<=200005;i++){ //修改值
if(a[i]>=k){
a[i]=1;
}else{
a[i]=0;
}
}
for(int i=1;i<=200005;i++){ //前缀和
a[i]+=a[i-1];
}
while(q–){
int x,y,ans=0;
scanf("%d%d",&x,&y);
cout<<a[y]-a[x-1]<<endl;
}
}

标签：200000,coffee,前缀,int,Coffee,admissible,Karen,between
来源： https://blog.csdn.net/weixin_52115456/article/details/120626442

本站声明： 1. iCode9 技术分享网（下文简称本站）提供的所有内容，仅供技术学习、探讨和分享；
2. 关于本站的所有留言、评论、转载及引用，纯属内容发起人的个人观点，与本站观点和立场无关；
3. 关于本站的所有言论和文字，纯属内容发起人的个人观点，与本站观点和立场无关；
4. 本站文章均是网友提供，不完全保证技术分享内容的完整性、准确性、时效性、风险性和版权归属；如您发现该文章侵犯了您的权益，可联系我们第一时间进行删除；
5. 本站为非盈利性的个人网站，所有内容不会用来进行牟利，也不会利用任何形式的广告来间接获益，纯粹是为了广大技术爱好者提供技术内容和技术思想的分享性交流网站。