题目链接

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题目大意：就是给你n个区间，从中选出k个区间，这k个区间共同覆盖了同一个点，问有多少种选法？结果mod 998244353

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解题思路：1.首先我们可以这么想：我们把区间左端点赋值为1，右端点赋值为-1，那么我们就可以像扫描线一样那么将左右端点分开存储，先按照第一关键排序然后如果第一关键字相同，那么就入度的点先进来。
2.当有一个点进来那么就从前面还存在的区间里面挑出k-1,一个点和它匹配那么不断的跟新答案就可以了

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代码：

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#include <iostream>
#include <cstdio>
#include <stack>
#include <sstream>
#include <limits.h>
#include <vector>
#include <map>
#include <cstring>
#include <deque>
#include <cmath>
#include <iomanip>
#include <unordered_map>
#include <queue>
#include <algorithm>
#include <set>
#define mid ((l + r) >> 1) 
#define Lson rt << 1, l , mid
#define Rson rt << 1|1, mid + 1, r
#define ms(a,al) memset(a,al,sizeof(a))
#define log2(a) log(a)/log(2)
#define _for(i,a,b) for( int i = (a); i < (b); ++i)
#define _rep(i,a,b) for( int i = (a); i <= (b); ++i)
#define for_(i,a,b) for( int i = (a); i >= (b); -- i)
#define rep_(i,a,b) for( int i = (a); i > (b); -- i)
#define lowbit(x) ((-x) & x)
#define IOS std::ios::sync_with_stdio(0); cin.tie(0); cout.tie(0)
#define INF 0x3f3f3f3f
#define LLF 0x3f3f3f3f3f3f3f3f
#define hash Hash
#define next Next
#define pb push_back
#define f first
#define s second
#define y1 Y
using namespace std;
const int N = 3e6 + 10, mod = 998244353;
const int maxn = 3e6;
const long double eps = 1e-5;
const int EPS = 500 * 500;
typedef long long ll;
typedef unsigned long long ull;
typedef pair<ll,ll> PII;
typedef pair<int,int> PLL;
typedef pair<double,double> PDD;
template<typename T> void read(T &x)
{x = 0;char ch = getchar();ll f = 1;while(!isdigit(ch)){if(ch == '-')f*=-1;ch=getchar();}while(isdigit(ch)){x = x*10+ch-48;ch=getchar();}x*=f;
}
template<typename T, typename... Args> void read(T &first, Args& ... args) 
{read(first);read(args...);
}
ll n ,k;
PII cl[maxn];
vector<ll> a;
ll tr[maxn + 5];
ll fac[maxn + 5];
void init()
{fac[0] = 1;for(ll i = 1; i < maxn; i ++)//阶乘表fac[i]=((ll)i * fac[i - 1]) % mod;
}ll quickpow(ll a,ll k)//求逆元
{ll res = 1;while(k){if(k & 1) res = (res * a) % mod;a = (a * a) % mod;k >>= 1;}return res;
}
ll C_n_m(ll n,ll m)//求C(n,m);
{ll ans = 1;ans = fac[n] * quickpow(fac[m] * fac[n - m] % mod,mod - 2) % mod;return ans;
}
ll lucas(ll n,ll m)
{ll ans=1;if(n < m) return 0;while(n && m && ans){ans=ans * C_n_m(n % mod,m % mod) % mod;n /= mod;m /= mod;}return ans;
}bool cmp(PII a, PII b)
{if(a.f == b.f) return a.s > b.s;else return a.f < b.f;
}int poi = 0;
int main()
{IOS;cin >> n >> k;init();for(int i = 0; i < n; ++ i){ll l, r;cin >> l >> r;cl[poi ++] = (PII){l,1};cl[poi ++] = (PII){r,-1};}sort(cl,cl+poi,cmp);ll ans = 0, tmp = 0;for(int i = 0; i < poi; ++ i){if(cl[i].s == 1){ll t = lucas(tmp % mod,k-1) % mod;ans = (ans + t) % mod;}tmp += cl[i].s;}cout << ans << endl;return 0;
}