Karen and Coffee 816B

Describe：
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n个区间，给定q个区间，输出每个区间内覆盖k次的元素个数.

思路：两次前缀和即AC，这题有用lazy-tag思想，对于给定的n个区间，左区间L标1，右区间R+1标-1. 一次操作前缀和得出每个元素被覆盖的次数，一次操作前缀和统计区间内覆盖k次以上元素的个数.

Coding

#include <iostream>#include <map>#include <cstdio>#include <cmath>#include <cstring>#define N 100010using namespace std;map<long long ,int > mp;int a[N],b[N];long long dex , x ,k ,n,p,ans;int main(){    cin >> n >> k;    for(int i = 0 ; i < n ;i ++){        scanf("%d",&a[i]);    }    p = 1;    while(abs(p) < 1e15){        mp.clear();        dex = 0 ;        mp[0] = 1;        for(int i = 0 ; i < n ;i++){            dex += a[i];            ans += mp[dex - p];            mp[dex]++;        }        p *= k;        if(p == 1)            break;    }    cout << ans << endl;}