CF1153F
题意：
在一个长度为\(l\)的区间里面\(n\)次随机选出一个子区间。
求至少被这\(n\)个子区间中\(k\)个覆盖的区间长度和的期望。
\(1\le k\le n\le 2000,1\le l\le 10^9\)
考虑长度和的期望，相当于是随机一个点在区间内的概率\(\times l\)
所以可以是按顺序随机\(2n+1\)个点\(a_{1,2\dots 2n+1}\)，然后\(a_{2k-1,2k}\)形成一个区间，求\(a_{2n+1}\)在至少\(k\)个区间中的概率\(\times l\)。
随机\(2n+1\)个点也相当于随机一个\(2n+1\)的排列。
发现\(2k-1,2k\)的相对顺序对答案不影响，并且\(2i-1,2i\)和\(2j-1,2j\)两对的顺序也对答案不影响。
考虑枚举\(2n\)个数的配对顺序，枚举\(2n+1\)的位置
设\(f_{i,j}\)表示枚举了前\(i\)位，当前有\(j\)个左端点没匹配到右端点的方案数。
那么总方案数是\(f_{2n,0}(2n+1)\)。
考虑枚举\(2n+1\)的位置\(i\)以及\(2n+1\)被覆盖的次数\(j\)，那么可行方案数是\(\sum_i \sum_{j\ge k}f_{i,j}f_{2n-i,j}\)。
然后做完了。
/*
Author: QAQ-Automaton
LANG: C++
PROG: F.cpp
Mail: cnyalilk@vip.qq.com
*/
#include<bits/stdc++.h>
#define int long long
#define debug(...) fprintf(stderr,__VA_ARGS__)
#define DEBUG printf("Passing [%s] in LINE %d\n",__FUNCTION__,__LINE__)
#define Debug debug("Passing [%s] in LINE %d\n",__FUNCTION__,__LINE__)
#define all(x) x.begin(),x.end()
#define x first
#define y second
using namespace std;
typedef long long ll;
typedef pair<int,int> pii;
#define inf 0x3f3f3f3f3f3f3f3f3f3f3f3f3f3f3f3f
const double eps=1e-8;
const double pi=acos(-1.0);
template<class T>int chkmin(T &a,T b){return a>b?a=b,1:0;}
template<class T>int chkmax(T &a,T b){return a<b?a=b,1:0;}
template<class T>T sqr(T a){return a*a;}
template<class T>T mmin(T a,T b){return a<b?a:b;}
template<class T>T mmax(T a,T b){return a>b?a:b;}
template<class T>T aabs(T a){return a<0?-a:a;}
template<class T>int dcmp(T a,T b){return a>b;}
template<int *a>int cmp_a(int x,int y){return a[x]<a[y];}
#define min mmin
#define max mmax
#define abs aabs
namespace io {
const int SIZE = (1 << 21) + 1;
char ibuf[SIZE], *iS, *iT, obuf[SIZE], *oS = obuf, *oT = oS + SIZE - 1, c, qu[55]; int f, qr;
// getchar
#define gc() (iS == iT ? (iT = (iS = ibuf) + fread (ibuf, 1, SIZE, stdin), (iS == iT ? EOF : *iS ++)) : *iS ++)
// print the remaining part
inline void flush () {
fwrite (obuf, 1, oS - obuf, stdout);
oS = obuf;
}
// putchar
inline void putc (char x) {
*oS ++ = x;
if (oS == oT) flush ();
}
// input a signed integer
inline bool read (signed &x) {
for (f = 1, c = gc(); c < '0' || c > '9'; c = gc()) if (c == '-') f = -1;else if(c==EOF)return 0;
for (x = 0; c <= '9' && c >= '0'; c = gc()) x = x * 10 + (c & 15); x *= f;
return 1;
}
inline bool read (long long &x) {
for (f = 1, c = gc(); c < '0' || c > '9'; c = gc()) if (c == '-') f = -1;else if(c==EOF)return 0;
for (x = 0; c <= '9' && c >= '0'; c = gc()) x = x * 10 + (c & 15); x *= f;
return 1;
}
inline bool read (char &x) {
x=gc();
return x!=EOF;
}
inline bool read(char *x){
while((*x=gc())=='\n' || *x==' '||*x=='\r')if(*x==EOF)return 0;
while(!(*x=='\n'||*x==' '||*x=='\r'))*(++x)=gc();
*x=0;
return 1;
}
template<typename A,typename ...B>
inline bool read(A &x,B &...y){
return read(x)&&read(y...);
}
// print a signed integer
inline bool write (signed x) {
if (!x) putc ('0'); if (x < 0) putc ('-'), x = -x;
while (x) qu[++ qr] = x % 10 + '0', x /= 10;
while (qr) putc (qu[qr --]);
return 0;
}
inline bool write (long long x) {
if (!x) putc ('0'); if (x < 0) putc ('-'), x = -x;
while (x) qu[++ qr] = x % 10 + '0', x /= 10;
while (qr) putc (qu[qr --]);
return 0;
}
inline bool write (char x) {
putc(x);
return 0;
}
inline bool write(const char *x){
while(*x){putc(*x);++x;}
return 0;
}
inline bool write(char *x){
while(*x){putc(*x);++x;}
return 0;
}
template<typename A,typename ...B>
inline bool write(A x,B ...y){
return write(x)||write(y...);
}
//no need to call flush at the end manually!
struct Flusher_ {~Flusher_(){flush();}}io_flusher_;
}
using io :: read;
using io :: putc;
using io :: write;
const int p=998244353;
int f[4005][2005];
int fac[4005];
int fpm(int a,int b){
int c=1;
for(;b;b>>=1,a=a*a%p)if(b&1)c=c*a%p;
return c;
}
signed main(){
#ifdef QAQAutoMaton
freopen("F.in","r",stdin);
freopen("F.out","w",stdout);
#endif
int k,n,l;
read(n,k,l);
f[0][0]=1;
fac[0]=1;
for(int i=0;i<n+n;++i){
fac[i+1]=fac[i]*(i+1)%p;
for(int j=0;j<=n;++j){
f[i+1][j+1]=(f[i+1][j+1]+f[i][j])%p;
if(j){
f[i+1][j-1]=(f[i+1][j-1]+f[i][j]*j)%p;
}
}
}
int cnt=0;
for(int i=1;i<n+n;++i){
for(int j=k;j<=n;++j){
cnt=(cnt+f[i][j]*f[n+n-i][j]%p*fac[j])%p;
}
}
write(cnt*fpm((n+n+1),p-2)%p*fpm(f[n+n][0],p-2)%p*l%p);
return 0;
}