CF1153F Serval and Bonus Problem

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首先我们考虑长度为(1)的情况，最后把答案乘上(l)即可。
随机一条线段，(xin[0,1])不被覆盖的概率为(p(x)=2x(1-x))。
利用二项分布的理论，(xin[0,1])被覆盖至少(k)次的概率为(f(x)=sumlimits_{i=k}^n{nchoose i}p(x)^i(1-p(x))^{n-i})。
那么此时的答案就是(int_0^1f(x)mathrm dx)。

[egin{aligned} ans&=int_0^1f(x)mathrm dx\ &=int_0^1sumlimits_{i=k}^n{nchoose i}p(x)^i(1-p(x))^{n-i}mathrm dx\ &=sumlimits_{i=k}^n{nchoose i}int_0^1p(x)^i(1-p(x))^{n-i}mathrm dx\ &=sumlimits_{i=k}^n{nchoose i}int_0^1p(x)^isumlimits_{j=0}^{n-i}{n-ichoose j}(-p(x))^jmathrm dx\ &=sumlimits_{i=k}^n{nchoose i}sumlimits_{j=0}^{n-i}{n-ichoose j}int_0^1p(x)^i(-p(x))^jmathrm dx\ &=sumlimits_{i=k}^n{nchoose i}sumlimits_{j=0}^{n-i}{n-ichoose j}(-1)^j2^{i+j}int_0^1x^{i+j}(1-x)^{i+j}mathrm dx\ &=sumlimits_{i=k}^n{nchoose i}sumlimits_{j=0}^{n-i}{n-ichoose j}(-1)^j2^{i+j}frac{((i+j)!)^2}{(2i+2j+1)!}\ &=n!sumlimits_{l=k}^n2^lfrac{(l!)^2}{(2l+1)!(n-l)!}sumlimits_{i+j=l}frac1{i!}frac{(-1)^j}{j!} end{aligned} ]

#include<cstdio>
#include<algorithm>
const int N=4097,P=998244353;
int lim,rev[N],w[N],fac[N],ifac[N],f[N],g[N],h[N];
int read(){int x;scanf("%d",&x);return x;}
int inc(int a,int b){return a+=b-P,a+(a>>31&P);}
int dec(int a,int b){return a-=b,a+(a>>31&P);}
int mul(int a,int b){return 1ll*a*b%P;}
int pow(int a,int k){int r=1;for(;k;k>>=1,a=mul(a,a))if(k&1)r=mul(a,r);return r;}
void init(int n)
{
    int p=(lim=1<<(32-__builtin_clz(n)))>>1,g=pow(3,(P-1)/lim);
    w[p]=1,fac[0]=1;
    for(int i=1;i<lim;++i) rev[i]=(rev[i>>1]>>1)|(i&1? p:0);
    for(int i=p+1;i<lim;++i) w[i]=mul(w[i-1],g);
    for(int i=p-1;i;--i) w[i]=w[i<<1];
    for(int i=1;i<=n;++i) fac[i]=mul(fac[i-1],i);
    ifac[n]=pow(fac[n],P-2);
    for(int i=n;i;--i) ifac[i-1]=mul(ifac[i],i);
}
void NTT(int*a,int f)
{
    if(!~f) std::reverse(a+1,a+lim);
    for(int i=1;i<lim;++i) if(i<rev[i]) std::swap(a[i],a[rev[i]]);
    for(int i=1;i<lim;i<<=1) for(int j=0,d=i<<1;j<lim;j+=d) for(int k=0,x;k<i;++k) x=mul(a[i+j+k],w[i+k]),a[i+j+k]=dec(a[j+k],x),a[j+k]=inc(a[j+k],x);
    if(!~f) for(int i=0,x=P-(P-1)/lim;i<lim;++i) a[i]=mul(a[i],x);
}
int main()
{
    int n=read(),k=read(),l=read(),ans=0;
    init(2*n+1);
    for(int i=k;i<=n;++i) f[i]=ifac[i];
    for(int i=0;i<=n;++i) g[i]=i&1? P-ifac[i]:ifac[i];
    NTT(f,1),NTT(g,1);
    for(int i=0;i<lim;++i) f[i]=mul(f[i],g[i]);
    NTT(f,-1);
    for(int i=1,pw=2;i<=n;pw=inc(pw,pw),++i) h[i]=mul(mul(mul(mul(fac[i],fac[i]),pw),ifac[i*2+1]),ifac[n-i]);
    for(int i=k;i<=n;++i) ans=inc(ans,mul(f[i],h[i]));
    printf("%d",mul(mul(ans,fac[n]),l));
}