Codeforces816B Karen and Coffee 2017-06-27 15:18 39人阅读 评论(0) 收藏

B. Karen and Coffee

time limit per test

2.5 seconds

memory limit per test

512 megabytes

input

standard input

output

standard output

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

Note

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.

———————————————————————————————————

题目的意思是给出n个区间初始为0每个区间+1，对每次查询问每段区间数大于看的数量

思路：对操作区间每个数++肯定TLE，所以我可以开一个数组b[i]表示i到最后都加1

那么区间[l,r]+1就可以表示为b[l]++,b[r+1]--; 先记录所有操作，在同一处理

查询前搞波前缀和就好了

#include <iostream>
#include <cstdio>
#include <cstring>
#include <string>
#include <algorithm>
#include <vector>
#include <queue>
#include <stack>
#include <map>
#include <set>
#include <cmath>

using namespace std;

#define LL long long
const int inf=0x7fffffff;

int a[200005];
int b[200005];
int pre[200005];

int main()
{
    int n,m,k,l,r;
   memset(a,0,sizeof a);
   memset(pre,0,sizeof pre);
   memset(b,0,sizeof b);
   scanf("%d%d%d",&n,&k,&m);
   for(int i=0;i<n;i++)
   {
        scanf("%d%d",&l,&r);
        a[l]++;
        a[r+1]--;
   }
   for(int i=1;i<200002;i++)
   {
       b[i]=b[i-1]+a[i];
       pre[i]=pre[i-1];
       if(b[i]>=k)
        pre[i]++;
   }
   for(int i=0;i<m;i++)
   {
        scanf("%d%d",&l,&r);
        printf("%d\n",pre[r]-pre[l-1]);

   }

    return 0;
}