【Codeforces 1420 D】Rescue Nibel!，卢卡斯组合数，排序，枚举贡献

problem

D. Rescue Nibel!
time limit per test2 seconds
memory limit per test256 megabytes
inputstandard input
outputstandard output
Ori and Sein have overcome many difficult challenges. They finally lit the Shrouded Lantern and found Gumon Seal, the key to the Forlorn Ruins. When they tried to open the door to the ruins… nothing happened.

Ori was very surprised, but Sein gave the explanation quickly: clever Gumon decided to make an additional defence for the door.

There are n lamps with Spirit Tree’s light. Sein knows the time of turning on and off for the i-th lamp — li and ri respectively. To open the door you have to choose k lamps in such a way that there will be a moment of time when they all will be turned on.

While Sein decides which of the k lamps to pick, Ori is interested: how many ways there are to pick such k lamps that the door will open? It may happen that Sein may be wrong and there are no such k lamps. The answer might be large, so print it modulo 998244353.

Input
First line contains two integers n and k (1≤n≤3⋅105, 1≤k≤n) — total number of lamps and the number of lamps that must be turned on simultaneously.

Next n lines contain two integers li ans ri (1≤li≤ri≤109) — period of time when i-th lamp is turned on.

Output
Print one integer — the answer to the task modulo 998244353.

Examples
inputCopy
7 3
1 7
3 8
4 5
6 7
1 3
5 10
8 9
outputCopy
9
inputCopy
3 1
1 1
2 2
3 3
outputCopy
3
inputCopy
3 2
1 1
2 2
3 3
outputCopy
0
inputCopy
3 3
1 3
2 3
3 3
outputCopy
1
inputCopy
5 2
1 3
2 4
3 5
4 6
5 7
outputCopy
7
Note
In first test case there are nine sets of k lamps: (1,2,3), (1,2,4), (1,2,5), (1,2,6), (1,3,6), (1,4,6), (2,3,6), (2,4,6), (2,6,7).

In second test case k=1, so the answer is 3.

In third test case there are no such pairs of lamps.

In forth test case all lamps are turned on in a time 3, so the answer is 1.

In fifth test case there are seven sets of k lamps: (1,2), (1,3), (2,3), (2,4), (3,4), (3,5), (4,5).

solution

/*
题意：
+ 给出n盏灯和他们的打开关闭时间，选择k盏灯。
+ 求满足在某一时刻同时打开的这k盏灯的选择方案数。
思路：
+ 将所有灯按开始和结束时间从小到大排序，逐个枚举每盏灯
+ 考虑当前灯对答案的贡献，即记前面开始比他早的灯中跟他重合的灯的个数为x，贡献即为C(x,k-1)，他们与当前灯共同构成一种不重复的情况。可以用优先队列维护。
*/
#include<bits/stdc++.h>
using namespace std;
typedef long long LL;
const int maxn = 1e6+10;
const int mod = 998244353;

LL fac[maxn], inv[maxn];
LL mpow(LL a, LL x) {
	if(x==0)return 1;
	LL t = mpow(a, x>>1);
	if(x%2==0)return t*t%mod;
	return t*t%mod*a%mod;
}
LL init(){
	fac[0]=inv[0]=1;
	for(int i = 1; i < maxn; i++){
		fac[i]=fac[i-1]*i%mod; inv[i]=mpow(fac[i],mod-2);
	}return 0;
}
LL C(int x, int y) {
	if(y<0 || y>x)return 0;
	return fac[x]*inv[y]%mod*inv[x-y]%mod;
}

int main(){
	ios::sync_with_stdio(false);
	LL n, k;  cin>>n>>k;
	vector<pair<LL,LL> >a(n);
	for(auto &[l, r]: a)cin>>l>>r;
	sort(a.begin(),a.end());
	
	LL ans = 0;  init();
	priority_queue<LL, vector<LL>, greater<LL> >q;
	for(int i = 0; i < n; i++){
		auto [l,r] = a[i];
		while(q.size() && q.top()<l)q.pop();
		ans += C(q.size(), k-1);
		ans %= mod;
		q.push(r);
	}
	cout<<ans<<"\n";
	return 0;
}



  

 

  
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原文链接：gwj1314.blog.csdn.net/article/details/113924878