[CF816B] Karen and Coffee（前缀和，思维）

题目链接：http://codeforces.com/contest/816/problem/B

题意：一个大区间被n条线覆盖，接下来有q次询问，询问[l,r]内有多少个点被>=k个线覆盖。

区间两头左闭右开打标记，左边+1右边-1，维护这个标记的前缀和，再维护一个前缀和，找到上一个前缀和里>=k的点，更新到当前的前缀和里就是了。 

#include <algorithm>
#include <iostream>
#include <iomanip>
#include <cstring>
#include <climits>
#include <complex>
#include <cassert>
#include <cstdio>
#include <bitset>
#include <vector>
#include <deque>
#include <queue>
#include <stack>
#include <ctime>
#include <set>
#include <map>
#include <cmath>
using namespace std;
#define fr first
#define sc second
#define cl clear
#define BUG puts("here!!!")
#define W(a) while(a--)
#define pb(a) push_back(a)
#define Rint(a) scanf("%d", &a)
#define Rll(a) scanf("%lld", &a)
#define Rs(a) scanf("%s", a)
#define Cin(a) cin >> a
#define FRead() freopen("in", "r", stdin)
#define FWrite() freopen("out", "w", stdout)
#define Rep(i, len) for(int i = 0; i < (len); i++)
#define For(i, a, len) for(int i = (a); i < (len); i++)
#define Cls(a) memset((a), 0, sizeof(a))
#define Clr(a, x) memset((a), (x), sizeof(a))
#define Full(a) memset((a), 0x7f7f7f, sizeof(a))
#define lrt rt << 1
#define rrt rt << 1 | 1
#define pi 3.14159265359
#define RT return
#define lowbit(x) x & (-x)
#define onenum(x) __builtin_popcount(x)
typedef long long LL;
typedef long double LD;
typedef unsigned long long ULL;
typedef pair<int, int> pii;
typedef pair<string, int> psi;
typedef pair<LL, LL> pll;
typedef map<string, int> msi;
typedef vector<int> vi;
typedef vector<LL> vl;
typedef vector<vl> vvl;
typedef vector<bool> vb;

const int maxn = 200200;
int n, k, q;
int line[maxn];
int sum[maxn];

signed main() {
    // FRead();
    int x, y;
    while(~Rint(n)) {
        Rint(k); Rint(q);
        Cls(line); Cls(sum);
        Rep(i, n) {
            Rint(x); Rint(y);
            line[x]++; line[y+1]--;
        }
        For(i, 1, 200020) {
            line[i] += line[i-1];
            if(line[i] >= k) sum[i] = sum[i-1] + 1;
            else sum[i] = sum[i-1];
        }
        W(q) {
            Rint(x); Rint(y);
            printf("%d
", sum[y]-sum[x-1]);
        }
    }
    RT 0;
}